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Find the sum of the following Arithmetic Progression.
(i) 50, 46, 42,... To 10 terms
(ii) 1, 3, 4, 7, . . . 26 to 12 terms.
(iii) 3, 9/2, 6, 15/2,.... to 25 terms.
(iv) 41, 36, 31, . . . To 12 terms.
(v) a + b, a - b, a -3b,... To 22 terms.
(vi) (x - y)^{2 }, (x^{2}, y^{2}), (x + y)^{2}, . . . to n terms.
(viii) – 26, – 24, – 22, . . . to 36 terms.
In the given problem, we need to find the sum of terms for different A.P.
So, here we use the following formula for the sum of n terms of an A.P.,
S_{n} = n/2[2a + (n − 1)d]
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
Common difference of the A.P. (d) = a_{2} - a_{1} = 46 - 50 = – 4
Number of terms (n) = 10
First term for the given A.P. (a) = 50 So, using the formula we get,
S_{10} = 10/2[2(50) + (10 − 1)(−4)] = (5)
[100 + (9)(-4)] = (5)
[100 - 36] = (6)
[64] = 320
Therefore, the sum of first 10 terms of the given A.P. is 320.
Common difference of the A.P. (d) = a_{2} - a_{1} = 3 - 1 = 2
Number of terms (n) = 12 First term (a) = 1 So, using the formula we get,
S_{12 }= 12/2[2(1) + (12 − 1)(2)] = (6)
[2 + (11)(2)] = (6)
[2 + 22 ] = (6)
[24] = 144
Therefore, the sum of first 10 terms of the given A.P. is 144
Common difference here is (d):
Number of terms (n) = 25 First term of the A.P. (a) = 3 So, using the formula we get,
S_{25} = 25/2[2(3) + (25 − 1)(3/2)]
= (25/2)[6+(72/2)(3/2)]
= (25/2)[6+(72/2)] = (25/2)[6+36]
= (25/2)[42] = (25)(21) = 525
Therefore, the sum of first 12 terms for the given A.P. is 162.
(iv) 41, 36, 31, . . . to 12 terms.
Common Difference of the A.P. (d) = a_{2} - a_{1} = 36 - 41 = -5
Number of terms (n) = 12
First term for the given A.P. (a) = 41 So, using the formula we get,
S_{12 }= 12/2[2(41) + (12 − 1)(−5)]
= (6)[82 + (11)(-5)]
= (6)[82 - 55]
= (6)[27] =162
Therefore, the sum of first 12 terms for the given A.P. 162
(v) a + b, a - b, a -3b,... to 22 terms.
Common difference of the A.P. (d)
= a_{2} - a_{1} = (a - b) - (a + b)
= a - b - a - b = -2b
Number of terms (n) = 22
First term for the given A.P. (a) = a + b So, using the formula we get,
S_{22 }= 22/2[2(a + b) + (22 − 1)( −2b)] = (11)
[2(a + b) + (22 - 1)(-2b)]
= (11)[2a + 2b + (21)(-2b)]
= (11)[2a + 2b - 42b]
= (11)[2a - 40b] = 22a - 40b
Therefore, the sum of first 22 terms for the given A.P. is: 22a - 40b
Common difference of the A.P. (d) = a_{2} - a_{1} = (x^{2} - y^{2}) - (x - y)^{2} = x^{2} + y^{2} - (x^{2} + y^{2} - 2xy) = 2xy
Number of terms (n) = n
First term for the given A.P. (a) = (x - y)^{2}
So, using the formula, we get.
S_{n }= n/2[2(x − y)^{2} + (n − 1)2xy]
Now, taking 2 common from both the terms inside bracket, we get
= n/2(2)[(x − y)^{2 }+ (n − 1)xy]
= (n)[(x - y)^{2} + (n - 1 )xy]
Therefore, the sum of first n terms of the given A.P. is (n)[(x - y)^{2} + (n - 1)xy]
= a_{2} - a_{1}
So, using the formula we get,
On further solving, we get
Therefore, the sum of first n terms for the given A.P. is
= a_{2} - a_{1} = (-24) - (-26) = -24 + 26 = 2
Number of terms (n) = 36
First term for the given A.P. (a) = -26
S_{n} = (36/2)[2(−26) + (36 − 1)(2)] = (18)
[- 52 + (35)(2) = (18)
[- 52 + 70] = (18)
(18) = 324
Therefore, the sum of first 36 terms for the given A.P. is 324
Find the sum to n term of the A.P. 5, 2, –1, –4, –7, . . .
In the given problem, we need to find the sum of the n terms of the given A.P. 5, 2, -1,- 4, – 7,....
Where; a = first term for the given A.P. d = common difference of the given A.P. and n = number of terms For the given A.P. (5, 2, -1,- 4,- 7,...),
Common difference of the A.P. (d) = a_{2}- a_{1} = 2 - 5 = – 3
First term for the given A.P. (a) = 5
S_{n} = n/2[2(5) + (n − 1)(−3)]
= n/2[10 + (−3n + 3)]
= n/2[10 − 3n +3 ]
= n/2[13 − 3n]
Therefore, the sum of first n terms for the given A.P. is n/2[13 − 3n]
Find the sum of n terms of an A.P. whose nth terms is given by a_{n} = 5 - 6n.
Here, we are given an AP, whose nth term is given by the following expression, a_{n} = 5 - 6n
So, here we can find the sum of the n terms of the given A.P., using the formula,
S_{n} = (n/2)(a + l)
Where, a = the first term l = the last term So, for the given AP,
The first term (a) will be calculated using n = 1 in the given equation for nth term of A.P.
a = 5 - 6(1) = 5 – 6 = -1
Now, the last term (l) or the nth term is given a_{n} = 5 - 6n
So, on substituting the values in the formula for the sum of n terms of an AP.,
we get, S/n =(n/2)[(−1) + 5 − 6n]
= (n/2)[4 − 6n]
= (n/2)(2)[2 − 3n]
= (n)(2 - 3n)
Therefore, the sum of the n terms of the given A.P. is (n) (2 - 3n)
Find the sum of first 15 term of each of the following sequences having n^{th} term as:
(i) a_{n} = 3 + 4n
(ii) b_{n} = 5 + 2n
(iii) x_{n} = 6 - n
(iv) y_{n} = 9 – 5n
(i) Here, we are given an A.P. whose nth term is given by the following expression,
a_{n} = 3 + 4n
We need to find the sum of first 15 term & n ,
S_{n} = n/2(a + l)
The first term (a) will be calculated using n = 1 in the given equation for n^{th} term of A.P.
a = 3 + 4(1) = 3 + 4 =7
Now, the last term (l) or the nth term is given l = a_{n} = 3 + 4n
So, on substituting the values in the formula for the sum of n terms of an A.P,
we get, S_{n} = 15/2(7 + 3 + 4(15))
= 15/2(10 + 60)
= 15/2(70)
= (15)(35)
= 525
Therefore, the sum of the 15 terms of the given A.P. is S_{15} = 525
(ii) Here, we are given an A.P. whose nth term is given by the following expression,
b_{n} = 5 + 2n
We need to find the sum of first 15 term & n,
b = 5 + 2(1) = 5 + 2 = 7
Now, the last term (l) or the n^{th} term is given l = b_{n} = 5 + 2n
So, on substituting the values in the formula for the sum of n terms of an A.P, we get,
S_{n} = 15/2(7 + 5 + 2(15))
= 15/2(12 + 30)
= 15/2(42)
= (15)(21)
= 315
Therefore, the sum of the 15^{th} term of the given A.P. is 315
(iii) Here, we are given an A.P. whose nth term is given by the following expression, x_{n} = 6 - n
We need to find the sum of first 15 term & n , So, here we can find the sum of the n terms of the given A.P., using the formula, S_{n} = n/2(a + l)
b = 6 - 1 = 5
Now, the last term (l) or the n^{th} term is given l = x_{n} = 6 - n
S_{n} = 15/2((5) + 6 − (15))
= 15/2(11 − 15)
= 15/2(−4)
= (15)(-2)
= -30
Therefore, the sum of the 15 terms of the given A.P. is -30.
(iv) Here, we are given an A.P. whose nth term is given by the following expression, y_{n} = 9 - 5n
So, here we can find the sum of the n terms of the given A.P., using the formula, S_{n} = n/2(a + l)
The first term (a) will be calculated using n = 1 in the given equation for n^{th }term of A.P.
b = 9 – 5 (1) = 9 - 5 = 4
Now, the last term (l) or the n^{th} term is given l = b_{n} = 9 - 5n
S_{n} = 15/2((4) + 9 − 5(15))
= 15/2(13 − 75)
= 15/2(−62)
= (15)(-31)
= – 465
Therefore, the sum of the 15 terms of the given A.P. is – 465
Find the sum of first 20 terms the sequence whose n^{th} term is a_{n} = An + B.
Here, we are given an A.P. whose nth term is given by the following expression a_{n} = An + B
We need to find the sum of first 20 terms.
a = A(1) + B = A + B
Now, the last term (l) or the n^{th} term is given l = a_{n} = An + B
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,
S_{20} = 20/2((A + B) + A(20) + B)
= 10[21A + 2B]
= 210A + 20B
Therefore, the sum of the first 20 terms of the given A.P. is 210 A + 20B
Find the sum of first 25 terms of an A.P whose n^{th} term is given by a_{n} = 2 - 3n.
Here, we are given an A.P. whose nth term is given by the following expression, a_{n} = 2 – 3n
We need to find the sum of first 25 terms.
So, here we can find the sum of the n terms of the given AP., using the formula,
The first term (a) will be calculated using n= 1 in the given equation for nth term of A.P.
a =2 -3(1) = 2 – 3 = -1
Now, the last term (1) or the nth term is given l = a_{n} = 2 - 3n
So, on substituting the values in the formula for the sum of n terms of an AP., we get,
S_{25} = 25/2[(−1) + 2 − 3(25)]
= 25/2[1 − 75]
= (25)(-37)
= – 925
Therefore, the sum of the 25 terms of the given A.P. is – 925
Find the sum of first 25 terms of an A.P whose n^{th} term is given by a_{n} = 7 - 3n.
Here, we are given an AP. whose n^{th} term is given by the following expression, a_{n} = 7 - 3n.
So, here we can find the sum of the n terms of the given AP., using the formula, S_{n} = n/2(a + l)
a = 7 - 3 = 4
Now, the last term (l) or the n^{th} term is given l = a_{n} = 7 - 3n
S_{n} = 25/2[(4) + 7 − 3(25)]
= 25/2[11 − 75]
= 25/2[−64]
= (25)(−32)
= −800
Therefore, the sum of the 25 terms of the given A.P. is S_{n} = – 800
If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, . . ., is 116. Find the last term.
In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the last term for that A.P.
So here, let us first find the number of terms whose sum is 116.
For that, we will use the formula, S_{n} = n/2[2a + (n − 1)d]
Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms So for the given A.P.(25, 22, 19,...)
The first term (a) = 25
The sum of n terms S_{n} = 116
Common difference of the A.P.
(d) = a_{2} - a_{1}
= 22 - 25
= -3
⟹ 116 = n/2[2(25) + (n − 1)(−3)]
⟹ (n/2)[50 + (−3 + 3)]
⟹ (n/2)[53 − 3n]
⟹ 116 × 2
= 53n - 3n^{2}
So, we get the following quadratic equation,
3n^{2} - 53n + 232 = 0
On solving by splitting middle term, we get,
⟹ 3n^{2} - 24n - 29n + 232 = 0
⟹ 3n( n - 8 ) - 29 ( n - 8 ) = 0
⟹ (3n - 29)( n - 8 ) = 0 Further, 3n - 29 = 0
⟹ n = 293 Also, n - 8 = 0
⟹ n = 8
Since, n cannot be a fraction, so the number of terms is 8. So, the term is:
a_{8} = a_{1} + 7d = 25 + 7(-3) = 25 - 21 = 4
Therefore, the last term of the given A.P. such that the sum of the terms is 116 is 4.
(i) How many terms of the sequence 18, 16, 14.... should be taken so that their sum is 0 (Zero).
(ii) How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?
(iii) How many terms of the A.P. 9, 17, 25, . . . must be taken so that their sum is 636?
(iv) How many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693?
(v) How many terms of the A.P. is 27, 24, 21. . . should be taken that their sum is zero?
(i) AP. is 18, 16,14,... So here, let us find the number of terms whose sum is 0.
For that, we will use the formula,
where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms The first term (a) = 18 The sum of n terms (S_{n}) = 0 Common difference of the A.P.
(d) = a_{2} - a_{1} = 16 - 18 = – 2
So, on substituting the values in the formula for the sum of n terms of an AP., we get
⟹ 0 = n/2[2(18) + (n − 1)(−2)]
⟹ 0 = n/2[36 + (−2n + 2)]
⟹ 0 = n/2[38 − 2n] Further, n/2
⟹ n = 0 Or, 38 - 2n = 0
⟹ 2n = 38
⟹ n = 19
Since, the number of terms cannot be zero; the number of terms (n) is 19
(ii) Here, let us take the common difference as d. So, we are given, First term (a_{1}) = -14 Filth term (a_{5}) = 2 Sum of terms (S_{n}) = 40
Now, a_{5} = a_{1 }+ 4d
⟹ 2 = -14 + 4d
⟹ 2 + 14 = 4d
⟹ 4d = 16
⟹ d = 4
Further, let us find the number of terms whose sum is 40.
S_{n }= n/2[2a + (n − 1)d]
Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms
The first term (a_{1}) = -14
The sum of n terms (S_{n}) = 40
Common difference of the A.P. (d) = 4
⟹ 40 = n/2[2(−14) + (n − 1)(4)]
⟹ 40 = n/2[−28 + (4n − 4)]
⟹ 40 = n/2[−32 + 4n]
⟹ 40(2) = - 32n + 4n^{2}
4n^{2} - 32n - 80 = 0
⟹ n^{2} - 8n + 20 = 0
On solving by splitting the middle term, we get
4n^{2} - 10n + 2n + 20 = 0
⟹ n(n - 10) + 2( n - 10 ) = 0
⟹ (n + 2)(n - 10) = 0 Further, n + 2 = 0
⟹ n = -2 Or, n - 10 = 0
⟹ n = 10
Since the number of terms cannot be negative.
Therefore, the number of terms (n) is 10.
(iii) AP is 9, 17, 25,...
So here, let us find the number of terms whose sum is 636.
The first term (a) = 9 The sum of n terms (S_{n}) = 636
(d) = a_{2} - a_{1} = 17 - 9 = 8
⟹ 636 = n/2[2(9) + (n − 1)(8)]
⟹ 636 = n/2[18 + (8n − 8)]
⟹ 636(2) = (n)[10 + 8n]
⟹ 1271 = 10n + 8n^{2}
⟹ 8n^{2} + 10n - 1272 = 0
⟹ 4n^{2}+ 5n - 636 = 0
On solving by splitting the middle term, we get,
⟹ 4n^{2} - 48n + 53n - 636 = 0
⟹ 4n(n - 12) - 53(n - 12) = 0
⟹ (4n - 53)(n - 12) = 0 Further, 4n - 53 = 0 ⟹ n = 534 Or, n - 12 = 0 ⟹ n = 12
Since, the number of terms cannot be a fraction.
Therefore, the number of terms (n) is 12.
(iv) A.P. is 63, 60, 57,...
So here. let us find the number of terms whose sum is 693.
For that, we will use the formula.
The first term (a) = 63
The sum of n terms (S_{n}) = 693
(d) = a_{2} - a_{1} = 60 - 63 = –3
So, on substituting the values in the formula for the sum of n terms of an AP we get.
⟹ 693 = n/2[2(63) + (n − 1)(−3)]
⟹ 693 = n/2[163+(−3n + 3)]
⟹ 693 = n/2[129 − 3n]
⟹ 693(2) = 129n - 3n^{2}
So. we get the following quadratic equation.
⟹ 3n^{2} - 129n + 1386 = 0
⟹ n^{2} - 43n + 462
On solving by splitting the middle term, we get.
⟹ n^{2} - 22n - 21n + 462 = 0
⟹ n(n - 22) -21(n - 22) = 0
⟹ (n - 22) (n - 21) = 0 Further, n - 22 = 0 ⟹ n = 22 Or, n - 21 = 0
⟹ n = 21 Here, 22^{nd} term will be a_{22} = a_{1} + 21d = 63 + 21( -3 ) = 63 - 63 = 0
So, the sum of 22 as well as 21 terms is 693.
Therefore, the number of terms (n) is 21 or 22 (v) A.P. is 27, 24, 21. . .
So here. let us find the number of terms whose sum is 0. For that, we will use the formula.
The first term (a) = 27
The sum of n terms (S_{n}) = 0
(d) = a_{2} - a_{1} = 24 - 27 = -3
⟹ 0 = n/2[2(27) + (n − 1)( − 3)]
⟹ 0 = (n)[54 + (n - 1)(-3)]
⟹ 0 = (n)[54 - 3n + 3]
⟹ 0 = n [57 - 3n] Further we have, n = 0 Or, 57 - 3n = 0
⟹ 3n = 57
The number of terms cannot be zero,
Therefore, the numbers of terms (n) is 19.
Find the sum of the first
(i) 11 terms of the A.P. : 2, 6, 10, 14, . . .
(ii) 13 terms of the A.P. : -6, 0, 6, 12, . . .
(iii) 51 terms of the A.P. : whose second term is 2 and fourth term is 8.
In the given problem, we need to find the sum of terms for different arithmetic progressions.
(i) 2, 6, 10, 14,... to 11 terms.
(d) = a_{2} - a_{1} = 10 - 6 = 4
Number of terms (n) = 11
First term for the given A.P. (a) = 2
So, using the formula we get, S_{11} = 11/2[2(2) + (11 − 1)4]
= 11/2[2(2) + (10)4]
= 11/2[4 + 40]
= 11 × 22
= 242
Therefore, the sum of first 11 terms for the given A.P. is 242
(ii) – 6, 0, 6, 12, ... to 13 terms.
Common difference of the AR
(d) = a_{2} - a_{1} = 6 - 0 = 6
Number of terms (n) = 13
First term for the given AP (a) = -6
S_{13} = 13/2[2(− 6) + (13 –1)6]
= 13/2[(−12) + (12)6]
= 13/2[60] = 390
Therefore, the sum of first 13 terms for the given AR is 390
(iii) 51 terms of an AP whose a_{2} = 2 and a_{4} = 8
Now, a_{2} = a + d
2 = a + d ...(i)
Also, a_{4} = a + 3
8 = a + 3d ... (2)
Subtracting (1) from (2), we get
2d = 6
d = 3
Substituting d = 3 in (i), we get 2 = a + 3
⟹ a = -1
Number of terms
(n) = 51
First terms for the given A.P.(a) = -1
So, using the formula, we get
S_{n} = 51/2[2(−1) + (51 − 1)(3)]
= 51/2[−2 + 150]
= 51/2[158]
= 3774
Therefore, the sum of first 51 terms for the A.P. is 3774.
Find the sum of (i) First 15 multiples of 8 (ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6. (iii) all 3 - digit natural numbers which are divisible by 13.
Where: a = first term for the given A.P. d = common difference of the given A.P. n = number of terms (i) First 15 multiples of 8.
So, we know that the first multiple of 8 is 8 and the last multiple of 8 is 120.
Also, all these terms will form an A.P. with the common difference of 8. So here,
First term (a) = 8 Number of terms (n) = 15
Common difference (d) = 8
Now, using the formula for the sum of n terms, we get
Therefore, the sum of the first 15 multiples of 8 is 960 (ii)(a)
First 40 positive integers divisible by 3 So, we know that the first multiple of 3 is 3 and the last multiple of 3 is 120.
Also, all these terms will form an A.P. with the common difference of 3.
So here, First term (a) = 3
Number of terms (n) = 40
Common difference (d) = 3
S_{n} = 40/2[2(3) + (40 − 1)3]
= 20[6 + (39)3]
= 20(6 + 117)
= 20(123) = 2460
Therefore, the sum of first 40 multiples of 3 is 2460 (b)
First 40 positive integers divisible by 5 So, we know that the first multiple of 5 is 5 and the last multiple of 5 is 200.
Also, all these terms will form an A.P. with the common difference of 5.
So here, First term (a) = 5
Common difference (d) = 5
S_{n }= 40/2[2(5) + (40 − 1)5]
= 20[10 + (39)5]
= 20 (10 + 195)
= 20 (205) = 4100
Therefore, the sum of first 40 multiples of 5 is 4100 (c)
First 40 positive integers divisible by 6 So, we know that the first multiple of 6 is 6 and the last multiple of 6 is 240.
Also, all these terms will form an A.P. with the common difference of 6.
So here, First term (a) = 6
Common difference (d) = 6
S_{n} = 40/2[2(6) + (40 − 1)6]
= 20[12 + (39)6]
=20(12 + 234)
= 20(246) = 4920
Therefore, the sum of first 40 multiples of 6 is 4920
(ii) All 3 digit natural number which are divisible by 13 So, we know that the first 3 digit multiple of 13 is 104 and the last 3 digit multiple of 13 is 988.
Also, all these terms will form an AR with the common difference of 13.
So here, First term (a) = 104 Last term (l) = 988
Common difference (d) = 13
So, here the first step is to find the total number of terms.
Let us take the number of terms as n. Now, as we know, a_{n} = a + (n − 1)d
So, for the last term,
988 = 104 + (n - 1)13
⟹ 988 = 104 + 13n -13
⟹ 988 = 91 + 13n
⟹ 13n = 897
⟹ n = 69
S_{n} = 69/2[2(104) + (69 − 1)13]
= 69/2[208 + 884]
= 69/2[1092]
= 69(546)
= 37674
Therefore, the sum of all 3 digit multiples of 13 is 37674
Find the sum:
(i) 2 + 4 + 6 + . . . + 200
(ii) 3 + 11 + 19 + . . . + 803
(iii) (-5) + (-8) + (-11) + . . . + (- 230)
(iv) 1 + 3 + 5 + 7 + . . . + 199
(vi) 34 + 32 + 30 + . . . + 10
(vii) 25 + 28 + 31 + . . . + 100
S_{n} = n/2[2a + (n - 1)d] or S_{n} = n/2[a + l]
Where; a = first term of the given A.P. d = common difference of the given A.P. l = last term n = number of terms
(i) 2 + 4 + 6 + ... + 200
(d) = a_{2} - a_{1} = 6 - 4 = 2
So here, First term (a) = 2
Last term (l) = 200
Common difference (d) = 2
Let us take the number of terms as n. Now, as we know, a_{n }= a + (n - 1)d
200 = 2 + (n - 1)2
200 = 2 + 2n - 2
200 = 2n
Further simplifying, n = 100
Now using the formula for sum of n terms,
S_{100} = 100/2[a + l] = 50
[2 + 200] = 50 X 202 = 10100
Therefore, the sum of the A.P is 10100
(d) = a_{2} - a_{1} =19 - 11 = 8
So here, First term (a) = 3 Last term (l) = 803
Let us take the number of terms as n.
Now, as we know, a_{n }= a + (n - 1)d
So, for the last term, further simplifying,
803 = 3 + (n - 1)8
⟹ 803 = 3 + 8n - 8
⟹ 803 + 5 = 8n
⟹ 808 = 8n
⟹ n = 101
S_{101} = 101/2[a + l]
= 101/2[3 + 803]
= 101/2[806]
= 101(403)
= 40703
Therefore, the sum of the A.P. is 40703
(iii) (-5) + (-8) + (-11) + . . . + (-230)
(d) = a_{2} - a_{2} = -8 - (-5) = -8 + 5 = -3
So here, First term (a) = -5 Last term (l) = -230
Common difference (d) = -3
Now, as we know, a_{n} = a + (n - 1)d
-230 = -5 + (n - 1)(-3)
⟹ -230 = -5 -3n +3
⟹ -230 + 2 = -3n
⟹ -228 = -3n
⟹ n = 76
S_{76} = 76/2[a + l] = 38[(- 5) + (-230)] = 38(-235) = – 8930
Therefore, the sum of the A.P. is – 8930
(d) = a_{2} - a_{1} = 3 - 1 = 2
So here, First term (a) = 1 Last term (l) = 199
Common difference (d) = 2 So, here the first step is to find the total number of terms.
So, for the last term, 199 = 1 + (n - 1)2
⟹ 199 = 1 + 2n - 2
⟹ 199 + 1 = 2n
⟹ n = 100
[1 + 199] = 50
(200) = 10000
Therefore, the sum of the A.P. is 10000
Common difference of the A.P. (d) = a_{2} - a_{1}
So here, First term (a) = 7 Last term (l) = 184
Common difference (d) = 72
84 = 7 + (n − 1) 7/2
84 = 7 + 7n/2 − 7/2
84(2) = 7 + 7n
7n = 161
n = 23
Now, using the formula for sum of n terms, we get
S_{n} = 23/2[2(7) + (23 − 1)7/2]
= 23/2[14 + (22)7/2]
= 23/2[14 + 77]
= 23/2[91]
= 2093/2
Therefore, the sum of the A.P. is 2093/2
(d) = a_{2} - a_{1} = 32 -34 = -2
So here, First term (a) = 34
Last term (l) = 10
Common difference (d) = -2
Let us take the number of terms as n. Now, as we know, a_{n} = a +(n - 1)d
⟹ 10 = 34 + (n - 1)(-2)
⟹ 10 = 34 - 2n + 2
⟹ 10 = 36 - 2n
⟹ 10 - 36 = -2n
Further solving for n,
⟹ -2n = -26
⟹ n = 13
S_{n} = 13/2[a + l]
= 13/2[34 + 10]
= 13/2[44]
= 12(22) = 286
Therefore, the sum of the A.P. is 286
(d) = a_{2} - a_{1} = 28 - 25 = 3
So here, First term (a) = 25 Last term (l) = 100
100 = 25 + (n - 1)(3)
100 = 25 + 3n - 3
100 = 22 + 3n
100 - 22 = 3n
Further solving for n, 78 = 3n n = 26
S_{n} = 26/2[a + l] = 13
[25 + 100] = 13
(125) = 1625
Therefore, the sum of the given A.P. is 1625
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
In the given problem, we have the first and the last term of an A.P. along with the common difference of the AP Here, we need to find the number of terms of the AP and the sum of all the terms.
Here, The first term of the A.P (a) = 17
The last term of the A.P (l) = 350
The common difference of the A.P. = 9
Let the number of terms be n. So, as we know that, l = a + (n - 1)d we get, 350 = 17 + (n- 1) 9
⟹ 350 = 17 + 9n - 9
⟹ 350 = 8 + 9n
⟹ 350 - 8 = 9n
Further solving this, n = 38
Using the above values in the formula,
S_{n} = n/2[a + l]
⟹ 38/2(17 + 350)
⟹ 19 × 367
⟹ 6973
Therefore, the number of terms is (n) 38 and the sum (S_{n}) is 6973
The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.
In the given problem, let us take the first term as a and the common difference as d. Here, we are given that,
a_{3} = 7 .... (1)
a_{7} = 3a_{3} + 2 .... (2)
So, using (1) in (2), we get,
a_{7 }= 3(7) + 2
= 21 + 2 = 23 .... (3)
Also, we know,
a_{n} = a +(n - 1)d
For the 3th term (n = 3), a_{3} = a + (3 - 1)d
⟹ 7 = a + 2d (Using 1)
⟹ a = 7 - 2d .... (4)
Similarly, for the 7th term (n = 7), a_{7} = a + (7 - I) d 24 = a + 6d (Using 3)
a = 24 - 6d .... (5)
Subtracting (4) from (5), we get,
a - a = (23 - 6d) - (7 - 2d)
⟹ 0 = 23 - 6d - 7 + 2d
⟹ 0 = 16 - 4d
Now, to find a, we substitute the value of d in (4), a =7 - 2(4)
⟹ a = 7 - 8
a = -1
So, for the given A.P, we have d = 4 and a = -1
So, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an AP,
Where; a = first term for the given A.P. d= common difference of the given A.P. n= number of terms So, using the formula for n = 20, we get,
S_{20} = 20/2[2(−1) + (20 − 1)(4)]
= (10)[-2 + (19)(4)]
= (10)[-2 + 76]
= (10)[74] = 740
Therefore, the sum of first 20 terms for the given A.P. is S_{20} = 740
The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.
In the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the common difference of the A.P.
Here, The first term of the A.P (a) = 2
The last term of the A.P (I) = 50
Sum of all the terms S„ = 442
Let the common difference of the A.P. be d.
So, let us first find the number of the terms (n) using the formula,
442 = (n/2)(2 + 50)
⟹ 442 = (n/2)(52)
⟹ 26n = 442
⟹ n = 17
Now, to find the common difference of the A.P. we use the following formula, l = a + (n - 1)d
We get, 50 = 2 + (17 - 1)d
⟹ 50 = 2 + 16d
⟹ 16d = 48
⟹ d = 3
Therefore, the common difference of the A.P. is d = 3
If 12^{th} term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms?
In the given problem, we need to find the sum of first 10 terms of an A.P.
Let us take the first term a and the common difference as d Here, we are given that,
a_{12} = -13 S_{4} = 24
a_{n }= a + (n - 1)d
For the 12th term (n = 12)
a_{12} = a + (12 - 1)d - 13
= a + 11d
a = -13 - 11d .... (1)
So, as we know the formula for the sum of n terms of an A.P. is given by,
So, using the formula for n = 4, we get,
S_{4} = 4/2[2(a) + (4 − 1)d]
⟹ 24 = (2)[2a + (3)(d)]
⟹ 24 = 4a + 6d
⟹ 4a = 24 - 6d
Subtracting (1) from (2), we get.
On further simplifying for d, we get,
⟹ -19 × 2 = 19d
⟹ d = – 2
Now, we have to substitute the value of d in (1),
a = -13 - 11(-2)
a = -13 + 22
a = 9
Now, using the formula for the sum of n terms of an A.P., for n = 10 we have,
S_{10} = 10/2[2(9) + (10 − 1)(−2)]
= (5)[19 + (9)(-2)]
= (5)(18 - 18) = 0
Therefore, the sum of first 10 terms for the given A.P. is S_{10} = 0.
Find the sum of first 22 terms of an A.P. in which d = 22 and a_{22} = 149.
In the given problem, we need to find the sum of first 22 terms of an A.P.
Let us take the first term as a. Here, we are given that,
a_{22} = 149 .... (1)
d = 22 .... (2)
a_{n} = a + ( n - 1) d
For the 22^{nd} term (n = 22),
a_{22} = a + (22 - 1)d
149 = a + (21) (22) (Using 1 and 2)
a = 149 - 462
a = - 313 .... (3)
where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms
So, using the formula for n = 22, we get,
S_{22} = 22/2[2(−313) + (22 − 1)(22)]
= (11)[ – 626 + 462]
= (11)[–164] = – 1804
Therefore, The sum of first 22 terms for the given A.P. is S_{22} = -1804
In an A.P., if the first tern is 22, the common difference is – 4 and the sum to n terms is 64, find n.
In the given problem, we need to find the number of terms of an A.P.
Let us take the number of terms as n. Here, we are given that, a = 22 d = – 4 S„ = 64
⟹ S_{n} = n/2[2(22) + (n − 1)(−4)]
⟹ 64 = n/2[2(22) + (n − 1)(−4)]
⟹ 64(2) = n(48 - 4n)
⟹ 128 = 48n - 4n^{2}
Further rearranging the terms, we get a quadratic equation,
4n^{2} - 48n + 128 = 0
On taking 4 common, we get,
n^{2} - 12n + 32 = 0
Further, on solving the equation for n by splitting the middle term, we get,
n^{2} - 8n - 4n + 32 = 0
n ( n - 8 ) - 4 ( n - 8 ) = 0
(n - 8) (n - 4) = 0
So, we get n - 8 = 0
Also, n - 4 = 0
⟹ n = 4
Therefore, n = 4 or 8.
In an A.P., if the 5^{th} and 12^{th} terms are 30 and 65 respectively, what is the sum of first 20 terms?
In the given problem, let us take the first term as a and the common difference d Here, we are given that,
a_{5} = 30 .... (1)
a_{12} = 65 .... (2)
a_{n} = a + (n - 1)d
For the 5th term (n = 5),
a_{5} = a + (5 - 1)d
30 = a + 4d (Using 1)
a = 30 - 4d .... (3)
Similarly, for the 12th term (n = 12),
a_{12} = a + (12 - 1) d
65 = a + 11d (Using 2)
a = 65 - 11d .... (4)
Subtracting (3) from (4), we get,
a - a = (65 - 11d) - (30 - 4d)
0 = 65 - 11d - 30 + 4d
0 = 35 - 7d
7d = 35
d = 5
Now, to find a, we substitute the value of d in (4).
a = 30 - 4(5)
a = 30 - 20
a = 10
So, for the given A.P.
d = 5 and a = 10
So, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,
Where;
a = first term of the given A.P.
So, using the formula for n = 20, we get
S_{20} = 20/2[2(10) + (20 − 1)(5)]
= (10)[20 + (19)(5)]
= (10)[20 + 95]
= (10)[115]
= 1150
Therefore, the sum of first 20 terms for the given A.P. is 1150
Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.
a_{2} = 14 .... (1)
a_{3} = 18 .... (2)
For the 2nd term (n = 2),
⟹ a_{2} = a + (2 - 1)d
⟹ 14 = a + d (Using 1)
⟹ a = 14 - d .... (3)
Similarly, for the 3^{rd} term (n = 3),
⟹ a_{3} = a + (3 - 1)d
⟹ 18 = a + 2d (Using 2)
⟹ a = 18 - 2d .... (4)
a - a = (18 - 2d) - (14 - d)
0 = 18 - 2d - 14 + d
0 = 4 - d
d = 4
Now, to find a, we substitute the value of d in (4),
a = 14 - 4
d = 4 and a = 10
So, to find the sum of first 51 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,
S_{n} = n/2(2a + (n − 1)d)
Where,
a = the first term of the A.P.
d = common difference of the A.P.
n = number of terms So, using the formula for
n = 51,
we get
S_{51} = 51/2[2(10) + (51 - 1)(4)]
= 51/2[20 + (40)4]
= 51/2[220]
= 51(110)
= 5610
Therefore, the sum of the first 51 terms of the given A.P. is 5610
The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Let a be the first term and d be the common difference. We know that, sum of first n terms is
Where, a = the first term of the A.P.
n = number of terms Also, nth term = a_{n} = a + (n - 1)
According to the question, First term (a) = 5, last term (a_{n}) = 45 and sum of n terms (S_{n}) = 400
Now,
⟹ 45 = 5 + (n - 1)d
⟹ 40 = nd - d
⟹ nd - d = 40 .... (1)
Also,
S_{n }= n/2(2(a) + (n − 1)d)
400 = n/2(2(5) + (n − 1)d)
800 = n (10 + nd - d)
800 = n (10 + 40) from (1)
n = 16 .... (2)
On substituting (2) in (1), we get
nd - d = 40
16d - d = 40
15d = 40
d = 8/3
Thus, common difference of the given A.P. is 83.
In an A.P. the first term is 8, n^{th} term is 33 and the sum of first n term is 123. Find n and the d, the common difference.
In the given problem, we have the first and the nth term of an A.P. along with the sum of the n terms of A.P.
Here, we need to find the number of terms and the common difference of the A.P
Here, The first term of the A.P (a) = 8
The nth term of the A.P (l)= 33 Sum of all the terms S_{n} = 123
123 = (n/2)(8 + 33)
123 = (n/2)(41)
n = 246/41
n = 6
Now, to find the common difference of the A.P. we use the following formula,
l = a + (n - 1)d we get
33 = 8 + (6 - 1)d
33 = 8 + 5d
5d = 25
Therefore, the number of terms is n = 6 and the common difference of the A.P. is
d = 5.
In an A.P. the first term is 22, n^{th} term is -11 and the sum of first n term is 66. Find n and the d, the common difference.
Here, we need to find the number of terms and the common difference of the A.P.
Here, The first term of the A.P (a) = 22
The nth term of the A.P (l) = -11
Sum of all the terms S = 66
66 = (n/2)[22 + (−11)]
66 = (n/2)[22 − 11]
(66)(2) = n(11)
6 × 2 = n
n = 12
l = a + (n - 1)d we get,
- 11 = 22 + (12 - 1 )d -11 = 22 + 11d
11d = – 33
d = – 3
Therefore, the number of terms is n = 12 and the common difference d = -3
The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find the common difference.
Let a be the first term and d be the common difference.
We know that, sum of first n terms is:
Also, n^{th }term (a_{n}) = a + (n - 1)d
According to question, first term (a) = 7 last term (a_{n}) = 49 and sum of n terms (S_{n}) = 420
⟹ 49 = 7 + (n - 1)d
⟹ 43 = nd - d
⟹ nd - d = 42 ..... (1)
S_{n} = n/2(2(7) + (n − 1)d)
⟹ 840 = n[14 + nd - d]
⟹ 840 = n[14 + 42] [from (1)]
⟹ 840 = 54n
⟹ n = 15 .... (2)
on substituting (2) in (1), we get
nd - d = 42
⟹ 15d - d = 42
⟹ 14d = 42
Thus, the common difference of the given A.P. is 3.
The sum of first q terms of an A.P. is 162. The ratio of its 6^{th} term to its 13^{th} term is 1: 2. Find the first and 15^{th} term of the A.P.
Let a be the first term and d be the common difference. We know that, sum of first n terms is:
Also, nth term = a_{n} = a + (n - 1)d
According to the question,
S_{q} = 162 and a_{6 }: a_{13} = 1 : 2
2a_{6 }= a_{13}
⟹ 2 [a + (6 - 1d)] = a + (13 - 1)d
⟹ 2a + 10d = a + 12d
⟹ a = 2d .... (1)
Also, S_{9} = 162
⟹ S_{9} = 9/2(2a + (9 − 1)d)
⟹ 162 = 9/2(2a + 8d)
⟹ 162 × 2 = 9[4d + 8d] [from (1)]
⟹ 324 = 9 × 12d
⟹ a = 2d [from (1)]
⟹ a = 6
Thus, the first term of the A.P. is 6
Now, a_{15} = a + 14d = 6 + 14 × 3 = 6 + 42
a _{15} = 48
Therefore, 15^{th} term of the A.P. is 48
If the 10^{th} term of an A.P. is 21 and the sum of its first 10 terms is 120, find its n^{th} term.
S_{n} = n/2(2a + (n − 1)d) and n^{th} term is given by:
Now, given in question,
S_{10} = 120
⟹ 120 = 10/2(2a + (10 − 1)d)
⟹ 120 = 5(2a + 9d)
⟹ 24 = 2a + 9d .... (1)
Also, a_{10} = 21
⟹ 21 = a + (10 - 1)d
⟹ 21 = a + 9d .... (2)
Subtracting (2) from (1), we get
24 - 21 = 2a + 9d - a - 9d
a = 3 Putting a = 3 in equation (2), we have
3 + 9d = 21
9d = 18
d = 2
So, we have now first term = 3 common difference = 2
Therefore, the n^{th} term can be calculated by:
a_{n} = a + (n - 1)d = 3 + (n - 1)2
= 3 + 2n -2
= 2n + 1
Therefore, the n^{th} term of the A.P is (a_{n}) = 2n + 1
The sum of first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28^{th} term of this A.P.
We know that, sum of first n terms
It is given that sum of the first 7 terms of an A.P. is 63.
And sum of next 7 terms is 161. Sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms
= 63 + 161 = 224
S_{7} = 7/2(2a + (7 − 1)d)
⟹ 63(2) = 7(2a + 6d)
⟹ 9 × 2 = 2a + 6d
⟹ 2a + 6d = 18 . . . . (1)
S_{14} = 14/2(2a + (14 − 1)d)
⟹ 224 = 7(2a + 13d)
⟹ 32 = 2a + 13d .... (2)
On subtracting (1) from (2), we get
⟹ 13d - 6d = 32 - 18
⟹ 7d = 14
⟹ d = 2
From (1) 2a + 6(2) = 18
2a = 18 - 12
a = 3
Also, n^{th} term = a_{n} = a + (n - 1)d
⟹ a_{28 }= a + (28 - 1)d
= 3 + 27 (2)
= 3 + 54 = 57
Thus, the 28^{th} term is 57.
The sum of first seven terms of an A.P. is 182. If its 4^{th} and 17^{th} terms are in ratio 1: 5, find the A.P.
In the given problem, let us take the first term as a and the common difference as d. Here, we are given that, S_{17} = 182 We know that, sum of first term is:
S_{n }= n/2(2a + (n − 1)d)
So, from question
182 × 2 = 7(2a + 6d)
364 = 14a + 42d
26 = a + 3d
a = 26 - 3d ... (1)
Also, we are given that 4^{th} term and 17^{th} term are in a ratio 0f 1 : 5 Therefore,
⟹ 5(a_{4}) = 1(a_{17})
⟹ 5 (a + 3d) = 1 (a + 16d)
⟹ 5a + 15d = a + 16d
⟹ 4a = d .... (2)
⟹ 4 ( 26 - 3d ) = d
⟹ 104 - 12d = d
⟹ 104 = 13d
⟹ d = 8 from (2), we get
⟹ 4a = d
⟹ 4a = 8
⟹ a = 2
Thus we get, first term a = 2 and the common difference d = 8.The required A.P. is 2, 10, 18, 26, . ..
In an A.P. the sum of first ten terms is -150 and the sum of its next 10 term is -550. Find the A.P.
Here, we are given S_{n}, = –150 and sum of the next ten terms is – 550.
Let us take the first term of the A.P. as a and the common difference as d. So, let us first find S_{10}.
For the sum of first 10 terms of this A.P, First term = a
Last term = a_{10} So, we know,
For the 10th term (n = 10),
a_{n} = a + (10 - 1)d = a + 9d
Where, a = the first term l = the last term So, for the given A.P,
S_{10} = (10/2)(a + a + 9d) - 150
= 5(2a + 9d) - 150
= 10a + 45d
Similarly, for the sum of next 10 terms (S_{10}),
First term = a_{11} Last term = a_{20}
For the 11th term (n = 11),
a_{11} = a+ (11 - 1)d
= a + 10d
For the 20th term (n = 20),
a_{20} = a + (20 - 1) d
= a + 19d
So, for the given AP,
S_{10} = (10/2)(a + 10d + a + 19d) - 550
= 5(2a + 29d) - 550
= 10a + 145d
Now subtracting (1) from (2),
0 = -550 - 145d + 150 + 45d
0 = - 400 - 100d
100d = - 400
d = - 4
Substituting the value of d in (1)
So, the A.P. is 3, -1, – 5, – 9,. . . with a = 3, d = – 4
In an A.P. , the first term is 2, the last term is 29 and the sum of the terms is 155, find the common difference of the A.P.
The last term of the AP (l) = 29
Sum of all the terms (S_{n}) = 155
155 = n/2(2 + 29)
155(2) = n(31)
31n = 310
n = 10
l = a + (n - 1)d
We get,
29 = 2 + (10 - 1)d
29 = 2 + (9)d
29 - 2 = 9d
9d = 27
Find the number of terms of the A.P. –12, –9, –6, . . . , 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.
First term, a_{1} = -12
Common difference,
d = a_{2} - a_{1} = – 9 – (- 12)
= - 9 + 12 = 3
n^{th} term = a_{n} = a + (n - 1)d
⟹ 21 = -12 + (n - 1)3
⟹ 21 = -12 + 3n - 3
⟹ 21 = 3n - 15
⟹ 36 = 3n
⟹ n = 12
Therefore, the number of terms is 12 Now, when 1 is added top each of the 12 terms, the sum will increase by 12.
So, the sum of all the terms of the A.P. thus obtained
⟹ S_{12 }+ 12 = 12/2[a + l] + 12
= 6[-12 + 21] + 12
= 6 × 9 + 12
= 66
Therefore, the sum after adding 1 to each of the term we get 66
The sum of first n terms of an A.P. is 3n^{2} + 6n. Find the n^{th} term of this A.P.
In the given problem, let us take the first term as a and the common difference as d. we know that n^{th} term is given by:
a_{n} = S_{n } - S_{n-1}
we have given here
S_{n} = 3n^{2} + 6n
So, using this to find the n^{th} term,
⟹ a_{n} = [3n^{2} + 6n] - [3(n - 1)^{2} + 6 (n - 1)]
= [3n^{2} + 6n] - [3(n^{2 }+ 1^{2} - 6n) + 6n - 6]
= 3n^{2} + 6n - 3n^{2} - 3 + 6n - 6n + 6 = 6n + 3
Therefore, the n^{th} term of this A.P. is 6n + 3
The sum of n terms of an A.P. is 5n - n^{2}. Find the n^{th} term of this A.P.
Let a be the first term and d be the common difference. We know that, sum of first n terms is :
S_{n} = n/2(2a + (n - 1)d)
It is given that sum of the first n terms of an A.P. is 5n - n^{2}.
First term = a = S_{1}= 5 (1) - (1)^{2} = 4.
Sum of first two terms = S_{2 }= 5(2) - (2)^{2} = 6.
Second term = S_{2} - S_{1} = 6 - 4 = 2.
Common difference = d = Second term - First term = 2 - 4 = – 2
Also, nth term = a_{n} = a + (n - 1) d
⟹ a_{n} = 4 + (n - 1)(-2)
⟹ a_{n} = 4 - 2n + 2
⟹ a_{n} = 6 - 2n
Thus, nth term of this A.P. is 6 - 2n.
The sum of first n terms of an A.P. is 3n^{2}+ 4n. Find the 25^{th} term of this A.P.
In the given problem, we have sum of n terms as S_{n} = 5n^{2} + 3n we know,
a_{n} = S_{n} - S_{n-1}
_{}We have to find out 25^{th} term, so n = 25
⟹ a_{25} = S_{25} - S_{24}
= [3(25)^{2} + 4 (25)] - [3 (24)^{2} + 4(24)]
= (3 × 625 + 100) - (3 × 576 + 96)
= 1975 - 1824 = 151
Therefore, its 25^{th} term is 151
The sum of first n terms of an A.P. is 5n^{2} + 3n. If its m^{th} term is 168, find the value of m. Also find the 20^{th} term of this A.P.
Here, we are given the Sum of the A.P. as
S_{n} = 5n^{2} + 3n. and its m^{th} term is a_{m }= 168
Let us assume its first term as a, and the common difference as d
We know, a_{n} = S_{n} - S_{n-1} So, here
⟹ a_{n} = (5n^{2} + 3n) - [5(n - 1)^{2} + 3 (n - 1)]
= 5n^{2} + 3n - [5(n^{2} + 1 - 2n) + 3n - 3]
= 5n^{2} + 3n - 5n^{2} -5 + 10n - 3n + 3
= 10n - 2
We are given, a_{m} = 168 Putting m in place of n, we get
⟹ a_{m} = 10m - 2
⟹ 168 = 10m - 2
⟹ 10m = 170
⟹ m = 17 and a_{20} = S_{20} - S_{19}
= [5(20)^{2} + 3 (20)] - [5 (19)^{2} + 3(19)]
= [2000 + 60] - [1805 + 57]
= 2060 - 1862 = 198
Therefore, in the given A.P. m = 17 and the 20^{th} term is a_{20 } = 198
If the sum of first n terms of an A.P. is 4n - n^{2}, what is the first term? What is the sum of first two terms? What is the second term? Similarly find the third, the tenth and the n^{th }term.
In the given problem, the sum of n terms of an A.P. is given by the expression,
S„ = 4n - n^{2}
So here, we can find the first term by substituting n = 1, S„ = 4n - n^{2} = 4(1) – 1^{2} = 4 - 1 = 3
Similarly, the sum of first two terms can be given by, S_{2} = 4(2) - (2)2 = 8 - 4 = 4
Now, as we know, a_{n} = S_{n} - S_{n-1} _{}So, a_{2} = S_{2} - S_{1} = 4 - 3 = 1
Now, using the same method we have to find the third, tenth and nth term of the A.P.
So, for the third term,
a_{3} = S_{3} - S_{2} = [4(3) - (3)^{2}] - [4(2) - (2)^{2}] = (12 - 9) - (8 - 4) = 3 - 4 = -1
Also, for the tenth term. a_{10 } = S_{10} - S_{9}
= [44(10) - (10)^{2}] - [4(9) - (9)^{2}]
= (40 - 100) - (36 - 81)
= - 60 + 45 = -15
So, for the nth term,
a_{n} = S_{n} - S_{n-1}
= [4(n) - (12)2] - [4(n - 1) - (n - 1)2]
= (4n - n^{2}) - (4n - 4 - n^{2} - 1 + 2n)
= 4n - n2 - 4n + 4 + n^{2} + 1 - 2n = 5 - 2n
Therefore, a = 3, S_{2} = 4, a_{2} = 1, a_{3} = -1, a_{10} = -15
If the sum of first n terms of an A.P. is 1/2(3n^{2} + 7n) , then find its n^{th} term. Hence write the 20^{th} term.
It is given that the sum of the first n terms of an A.P. is:
1/2(3n^{2} + 7n)
Therefore, first term (a)
= S_{1} = 1/2(3(1)^{2} + 7(1))
= 1/2(3X1 + 7)
= 1/2(10) = 5
Sum of first two terms
= S_{2} = 1/2(3(2)^{2} + 7(2))
= 1/2(3X4 + 14)
= 1/2(26) = 13
Therefore, second term = S_{2} - S_{1} = 13 - 5 = 8
Common difference = d = second term - first term = 8 - 5 = 3
Also, n^{th} term of the A.P. is : a + (n - 1)d
= 5 + (n - 1)3 = 5 + 3n - 3 = 3n + 2
Thus, n^{th} term of this A.P. is 3n + 2.
Now, we have to find the 20^{th} term, so
Putting n = 2o in the above equation, we get a_{20} = 3 (20) + 2 = 60 + 2 = 62
Thus, 20^{th} term of this A.P. is 62.
In an A.P. the sum of first n terms is
Find its 25^{th} term.
Here the sum of first n terms is given by the expression,
We need to find the 25^{th} term of the A.P. So, we know that the n^{th} term of an A.P. is given by, a_{n} = S_{n} - S_{n-1 } So,
a_{25} = S_{25 }- S_{24 }.... (1)
So, using the expression for the sum of n terms, we find the sum of 25 terms (S_{25}) and the sum of 24 terms (S_{25}), we get,
Now, using the above values in (1), a_{21} = S_{25} - S_{24} = 1100 - 1020 = 80
Therefore, a_{25} = 80
Find the sum of all natural numbers between 1 and 100, which are divisible by 3.
In this problem, we need to find the sum of all the multiples of 3 lying between 1 and 100.
So, we know that the first multiple of 3 after 1 is 3 and the last multiple of 3 before 100 is 99.
So here, First term (a) = 3 Last term (l) = 99
So, for the last term, 99 = 3 + (n - 1)3
⟹ 99 = 3 + 3n - 3
⟹ 99 = 3n
Further simplifying, ⟹ n = 33
Now, using the formula for the sum of n terms, i.e.
S_{n} = n/2[2a + (n − 1)d] we get,
⟹ S_{33} = 33/2[2(3) + (33 − 1)3]
= S_{33} = 33/2[6 + (32)3]
= S_{33} = 33/2[6 + 96]
= 33 (51) = 1683
Therefore, the sum of all the multiples of 3 lying between 1 and 100 is S_{n} = 1683
Find the sum of all odd numbers between (i) 0 and 50 (ii) 100 and 200.
(i) In this problem, we need to find the sum of all odd numbers lying between 0 and 50.
So, we know that the first odd number after 0 is 1 and the last odd number before 50 is 49.
Also, all these terms will form an AP. with the common difference of 2.
So here, First term (a) = 1
Last term (0 = 49 Common difference (d) = 2
Now, as we know, a_{n} = a + (n -1)d
⟹ 49 = 1 + (n - 1)d
⟹ 49 = 1 + 2n - 2
⟹ 49 = 2n - 2
⟹ 49 + 1 = 2n
Further simplifying,
⟹ 50 = 2n
⟹ n = 25
Now, using the formula for the sum of n terms,
⟹ S_{n} = n/2[2a + (n − 1)d] for n = 25, we get
⟹ S_{25 }= 25/2[2(1) + (25 − 1)2]
= 25/2[2 + 24 × 2]
= 25 × 25 = 625
Therefore, the sum of all the odd numbers lying between 0 and 50 is 625.
(ii) In this problem, we need to find the sum of all odd numbers lying between 100 and 200. So, we know that the first odd number after 0 is 101 and the last odd number before 200 is 199. Also, all these terms will form an AR. with the common difference of 2. So here, First term (a) = 101 Last term (a_{n}) = 199 Common difference (d) = 2 So, here the first step is to find the total number of term. Let us take the number of terms as n. Now, as we know, a_{n} = a + (n - 1)d So, for the last term,
⟹ 199 = 101 + (n - 1)2
⟹ 199 = 101 + 2n - 2
⟹ 199 = 99 + 2n
⟹ 199 - 99 = 2n
⟹ 100 = 2n
⟹ n = 50
S_{n} = n/2[2(a) + (n − 1)d]
For n = 50, we get
⟹ S_{50} = 50/2[2(101) + (50 − 1)2]
= 25[202 + (49)2]
= 25(202 + 98) = 25
(300) = 7500
Therefore, the sum of all the odd numbers lying between 100 and 200 is 7500
Find the sum of all integers between 84 and 719, which are multiples of 5.
In this problem, we need to find the sum of all the multiples of 5 lying between 84 and 719.
So, we know that the first multiple of 5 after 84 is 85 and the last multiple of 5 before 719 is 715.
Also, all these terms will form an A.P. with the common difference of 5. So here, First term (a) = 85
Last term (l) = 715 Common difference (d) = 5
Now, as we know, a = a +(n -1)d
So, for the last term.
715 = 85 + (n - 1)5
715 = 85 + 5n - 5
715 = 80 + 5n
715 - 80 = 5n
635 = 5n
n = 127
For n = 127,
S_{127} = 127/2[2(85) + (127 − 1)5]
= 127/2[170 + 630]
Therefore, the sum of all the multiples of 5 lying between 84 and 719 is 50800.
Find the sum of all integer between 50 and 500, which are divisible by 7.
In this problem, we need to find the sum of all the multiples of 7 lying between 50 and 500. So, we know that the first multiple of 7 after 50 is 56 and the last multiple of 7 before 500 is 497.
Also, all these terms will form an A.P. with the common difference of 7.
So here, First term (a) = 56 Last term (l) = 497
Common difference (d) = 7 So, here the first step is to find the total number of terms.
Now, as we know, a_{n} = a + (n - 1)d So, for the last term. 497 = 56 + (n - 1)7
⟹ 497 = 56 + 7n - 7
⟹ 497 = 49 + 7n
⟹ 497 - 49 = 7n
448 = 7n
n = 64
S_{n} = n/2[2a + (n − 1)d] for n = 64, we get
S_{64} = 64/2[2(56) + (64 − 1)7]
= 32[112 + (63)7]
= 32[112 + 441]
= 32(553) = 17696
Therefore, the sum of all the multiples of 7 lying between 50 and 500 is 17696
Find the sum of all even integers between 101 and 999.
In this problem, we need to find the sum of all the even numbers lying between 101 and 999.
So, we know that the first even number after 101 is 102 and the last even number before 999 is 998.
Also, all these terms will form an A.P. with the common difference of 2.
So here, First term (a) = 102 Last term (l) = 998
Common difference (d) = 2 So, here the first step is to find the total number of terms. Let us take the number of terms as n.
⟹ 998 = 102 + (n - 1)2
⟹ 998 = 102 + 2n - 2
⟹ 998 = 100 + 2n
⟹ 998 - 100 = 2n
⟹ 898 = 2n
⟹ n = 449
S_{n} = n/2[2a + (n − 1)d] For n = 449, we get
S_{449} = 449/2[2(102) + (449 − 1)2]
= 449/2[204 + (448)2]
= 449/2[204 + 896]
= 449/2[1100] = 449
(550) = 246950
Therefore, the sum of all even numbers lying between 101 and 999 is 246950
Find the sum of all integers between 100 and 550, which are divisible by 9.
In this problem, we need to find the sum of all the multiples of 9 lying between 100 and 550.
So, we know that the first multiple of 9 after 100 is 108 and the last multiple of 9 before 550 is 549.
Also, all these terms will form an A.P. with the common difference of 9.
So here, First term (a) = 108
Last term (l) = 549
Common difference (d) = 9 So, here the first step is to find the total number of terms.
⟹ 549 = 108 + (n - 1)d
⟹ 549 = 108 + 9n - 9
⟹ 549 = 99 + 9n
⟹ 549 - 99 = 9n
Further simplifying
⟹ 9n = 450
S_{n} = n/2[2a + (n − 1)d] We get,
S_{n} = 50/2[2(108) + (50 − 1)9] = 25
[216 + (49)9] = 25
(216 + 441) = 25
(657) = 16425
Therefore, the sum of all the multiples of 9 lying between 100 and 550 is 16425
Let there be an A.P. with first term ‘a’, common difference ‘d’. If a_{n} denotes its n^{th }term and S_{n} the sum of first n terms, find.
(i) n and S_{n,} if a = 5 , d = 3 , and a_{n} = 50.
(ii) n and a, if a_{n} = 4 , d = 2 and S_{n} = -14.
(iii) d, if a = 3, n = 8 and S_{n} = 192.
(iv) a, if a_{n} = 28, S_{n} = 144 and n = 9.
(v) n and d, if a = 8, a_{n} = 62 and S_{n }= 120.
(vi) n and a_{n} , if a = 2, d = 8 and S_{n} = 90.
(i) Here, we have an A.P. whose nth term (a_{n}), first term (a) and common difference (d) are given.
We need to find the number of terms (n) and the sum of first n terms (S_{n}).
Here, First term (a) = 5 Last term (a_{n}) = 50
So here we will find the value of n using the formula, a_{n} = a + (n - 1)d
So, substituting the values in the above mentioned formula
⟹ 50 = 5 + (n - 1)3
⟹ 50 = 5 + 3n - 3
⟹ 50 = 2 + 3n
⟹ 3n = 50 - 2
Further simplifying for n, 3n = 48 n =16
Now, here we can find the sum of the n terms of the given A.P., using the formula,
Where, a = the first term l = the last term
So, for the given A.P, on substituting the values in the formula for the sum of n terms of an A.P., we get,
S_{16 }= (16/2)(5 + 50) = 8
(55) = 440
Therefore, for the given A.P. we have, n = 16 and S_{16} = 440
(ii) Here, we have an A.P. whose nth term (an), sum of first n terms (S0) and common difference (d) are given.
We need to find the number of terms (n) and the first term (a).
Here, Last term (l) = 4
Sum of n terms (S_{n}) = -14
⟹ 4 = a + (n - 1)2
⟹ 4 = a + 2n - 2
⟹ 4 + 2 = a + 2n
Now, here the sum of the n terms is given by the formula,
S/n = (n/2)(a + l)
Where, a = the first term l = the last term So, for the given A.P, on substituting the values in the formula for the sum of n terms of an A.P., we get,
⟹ −14 = n/2(a + 4)
⟹ 14(2) = n(a + 4)
⟹ n = −28a + 4 .... (2)
Equating (1) and (2), we get,
= – 28(2) 6a - a^{2} + 24 - 4a
= – 56 - a^{2} + 2a + 24 + 56 = 0
- a ^{2} + 2a + 80 = 0
a^{2} - 2a - 80 = 0
Further solving it for a by splitting the middle term,
a^{2} - 10a + 8a - 80 = 0
a(a - 10) + 8(a - 10) = 0
(a - 10)(a + 8) = 0
So, we get,
a - 10 = 0
a = 10 or, a + 8 = 10
a = – 8
Substituting, a = 10 in (1)
n = (-4)/2
n = -2
Here, we get n as negative, which is not possible. So, we take a = - 8
n = 14/2
n = 7
Therefore, for the given A.P. n = 7 and a = – 8
(iii) Here, we have an A.P. whose first term (a), sum of first n terms (S0) and the number of terms (n) are given. We need to find common difference (d).
Here, First term (a) = 3 Sum of n terms (S_{n}) = 192
Number of terms (n) = 8
So here we will find the value of n using the formula, a_{n} = a + (n -1) d
So, to find the common difference of this A.P., we use the following formula for the sum of n terms of an A.P
S_{n} = (n/2)[2a + (n − 1)d]
n = number of terms So, using the formula for n = 8, we get,
S_{8} = (8/2)[2(3) + (8 − 1)d]
192 = 4[6 + 7d]
192 = 24 + 28d
28d = 192 - 24
28d = 168
d = 6
Therefore, the common difference of the given A.P. is d = 6
(iv) Here, we have an A.P. whose nth term (an), sum of first n terms (S_{n}) and the number of terms (n) are given. We need to find first term (a).
Here, Last term (a_{9}) = 28
Sum of n terms (S_{n}) = 144
Number of terms (n) = 9 Now, a_{9} = a + 8d
28 = a + 8d .... (1)
Also, using the following formula for the sum of n terms of an AP
n = number of terms So, using the formula for n = 9, we get,
S_{9} = (9/2)[2a + (9 − 1)d]
144(2) = 9[2a + 8d]
288 = 18a + 72d .... (2)
Multiplying (1) by 9, we get 9a + 72d = 252 .... (3)
Further, subtracting (3) from (2), we get 9 a = 36 a = 4
Therefore, the first term of the given A.P. is a = 4
(v) Here, we have an A.P. whose nth term (an), sum of first n terms (S_{n}) and first term (a) are given.
We need to find the number of terms (n) and the common difference (d).
Here, First term (a) = 8
Last term (a_{n }) = 62 Sum of n terms (S_{n}) = 210
210 = n/2[8 + 62]
210(2) = n (70)
n = 420/70
Also, here we will find the value of d using the formula,
a_{n} = a + (n - 1) d
62 = 8 + (6 - 1)d
5d = 54
d = 545
Therefore, for the given A.P. n = 6 and d = 545
(vi) Here, we have an A.P. whose first term (a), common difference (d) and sum of first n terms are given.
We need to find the number of terms (n) and the nth term (a_{n}).
Here, First term (a) = 2
Sum of first nth terms (S_{n}) = 90
Common difference (d) = 8 So, to find the number of terms (n) of this A.P., we use the following formula for the sum of n terms of an A.P
d = 8, we get,
S_{n} = (n/2)[2(2) + (n – 1)8]
90 = (n/2)[4 + 8n - 8]
90(2) = n[8n - 4]
180 = 8n^{2} - 4n
Further solving the above quadratic equation,
8n^{2} - 4n - 180 = 0
2n^{2} - n - 45 = 0
Further solving for n, 2n^{2} - 10n + 9n - 45 = 0
2n(n - 5) + 9(n - 5) = 0
(2n - 9)(n - 5) = 0
Now, 2n + 9 = 0
n - 5 = 0
n = 5
Since, n cannot be a fraction.
Thus, n = 5 Also, we will find the value of n^{th} term (a_{n}), using the formula,
So, substituting the values in the above formula,
a_{n }= 2 + (5 - 1) 8
a_{n} = 2 + 4 (8)
a_{n} = 2 + 32
a_{n} = 34
Therefore, for the given A.P., n = 5 and a_{n} = 34
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Chapter 9: Arithmetic Progressions Exercise 9.2...