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Chapter 9: Arithmetic Progressions Exercise 9.5 Question: 1 Find the sum of the following arithmetic progressions: (i) 50, 46, 42, ... to 10 terms (ii) 1, 3, 5, 7, ... 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, (x2 + y2), (x + y)2, to 22 tams (viii) – 26, – 24, – 22, .... to 36 terms Solution: In an A.P let first term = a, common difference = d, and there are n terms. Then, sum of n terms is, (i) Given progression is, 50, 46, 42 to 10 term. First term (a) = 50 Common difference (d) = 46 - 50 = – 4 nth term = 10 = 5{100 - 9.4} = 5{100 - 36} = S × 64 ∴ S10 = 320 (ii) Given progression is, 1, 3, 5, 7, .....to 12 terms First term difference (d) = 3 - 1 = 2 nth term = 12 = 6 × {2 + 22} = 6.24 ∴ S12 = 144. (iii) Given expression is 3, 9/2, 6, 15/2, ... to 25 terms First term (a) = 3 Common difference (d) = 9/2 - 3 = 3/2 Sum of nth terms Sn, given n = 25 (iv) Given expression is, 41, 36, 31 to 12 terms. First term (a) = 41 Common difference (d) = 36 - 41 = - 5 Sum of nth terms Sn, given n = 12 (v) a + b, a – b, a - 3b to 22 terms First term (a) = a + b Common difference (d) = a - b - a - b = -2b Sum of nth terms Sn = n/2{2a(n - 1). d} Here n = 22 S22 = 22/2{2.(a + b) + (22 - 1). -2b} = 11{2(a + b) - 22b) = 11{2a - 20b} = 22a - 440b ∴S22 = 22a - 440b (vi) (x - y)2,(x2 + y2), (x + y)2,... to n terms First term (a) = (x - y)2 Common difference (d) = x2 + y2 - (x - y)2 = x2 + y2 - (x2 + y2 - 2xy) = x2 + y2 - x2 + y2 + 2xy = 2xy Sum of nth terms Sn = n/2{2a(n - 1). d} = n/2{2(x - y)2 + (n - 1). 2xy} = n{(x - y)2 + (n - 1)xy} ∴ Sn = n{(x — y)2 + (n — 1). xy) (viii) Given expression -26, - 24. -22, to 36 terms First term (a) = -26 Common difference (d) = -24 - (-26) = -24 + 26 = 2 Sum of n tams Sn = n/2{2a + (n - 1)d) Sum of n tams Sn = 36/2{2. -26 + (36 - 1)2} = 18[-52 + 70] = 18.18 = 324 ∴ Sn = 324 Question: 2 Find the sum to n tam of the A.P. 5, 2, –1, – 4, –7, ... Solution: Given AP is 5, 2, -1, -4, -7, ..... a = 5, d = 2 - 5 = -3 Sn = n/2{2a + (n - 1)d} = n/2{2.5 + (n - 1) - 3} = n/2{10 - 3(n - 1)} = n/2{13 - 3n) ∴ Sn = n/2(13 - 3n) Question: 3 Find the sum of n terms of an A.P. whose the terms is given by an = 5 - 6n. Solution: Given nth term an = 5 - 6n Put n = 1, a1 = 5 - 6.1 = -1 We know, first term (a1) = -1 Last term (an) = 5 - 6n = 1 Then Sn = n/2(-1 + 5 - 6n) = n/2(4 - 6n) = n/2(2 - 3n) Question: 4 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. Solution: Given AP is 25, 22, 19, .......... First term (a) = 25, d = 22 - 25 = -3. Given, Sn = n/2(2a + (n - 1)d) 116 = n/2(2 × 25 + (n - 1) - 3) 232 = n(50 - 3(n - 1)) 232 = n(53 - 3n) 232 = 53n - 3n2 3n2 - 53n + 232 = 0 (3n - 29)(n - 8) = 0 ⟹ ag = 25 + (8 - 1) – 3 ∴ n = 8, ag = 4 = 25 - 21 = 4 Question: 5 (i) How many terms of the sequence 18, 16, 14, ... should be taken so that their (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? Solution: (i) Given sequence, 18, 16, 14, ... a = 18, d = 16 - 18 = -2. Let, sum of n terms in the sequence is zero Sn = 0 n/2(2a + (n. - 1)d) = 0 n/2(2.18 + (n - 1) - 2) = 0 n(18 - (n - 1)) = 0 n(19 - n) = 0 n = 0 or n = 19 (ii) ∵ n = 0 is not possible. Therefore, sum of 19 numbers in the sequence is zero. Given, a = -14, a 5 = 2 a+ (5 - 1)d = 2 -14 + 4d = 2 4d = 16 ⟹ d = 4 Sequence is -14, -10, -6, -2, 2, ........... Given Sn = 40 40 = n/2{2(-14) + (n - 1)4} 80 = n(-28 + 4n - 4) 80 = n(-32 + 4n) 4(20) = 4n(-8 + n) n2 - 8n - 20 = 0 (n - 10)(n + 2) = 0 n = 10 or n = -2 Sum of 10 numbers is 40 (Since -2 is not a natural number) (iii) Given AP 9. 17, 25, .................. a = 9, d = 17 - 9 = 8, and Sn = 636 636 = n/2(2.9 + (n - 1)8) 1272 = n(18 - 8 + 8n) 1272 = n(10 + 8n) 2 × 636 = 2n(5 + 4n) 636 = 5n + 4n2 4n2 + 5n - 636 = 0 (4n + 53)(n - 12) = 0 ∴ n = 12 (Since n (-53)/4 is not a natural number) Therefore, value of n is 12. (iv) Given AP, 63, 60, 57, ............. a = 63, d = 60 - 63 = -3 Sn = 693 Sn = n/2(2a + (n - 1)d) 693 = n/2(2.63 + (n. - 1) - 3) 1386 = n(126 - 3n + 3) 1386 = (129 - 3n)n 3n2 - 129n + 1386 = 0 n2 - 43n + 462 = 0 n = 21, 22 ∴ Sum of 71 or 77 tern is 693
Find the sum of the following arithmetic progressions:
(i) 50, 46, 42, ... to 10 terms
(ii) 1, 3, 5, 7, ... 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, (x2 + y2), (x + y)2, to 22 tams
(viii) – 26, – 24, – 22, .... to 36 terms
In an A.P let first term = a, common difference = d, and there are n terms.
Then, sum of n terms is,
(i) Given progression is,
50, 46, 42 to 10 term.
First term (a) = 50
Common difference (d) = 46 - 50 = – 4
nth term = 10
= 5{100 - 9.4}
= 5{100 - 36}
= S × 64
∴ S10 = 320
(ii) Given progression is, 1, 3, 5, 7, .....to 12 terms
First term difference (d) = 3 - 1 = 2 nth term = 12
= 6 × {2 + 22} = 6.24
∴ S12 = 144.
(iii) Given expression is 3, 9/2, 6, 15/2, ... to 25 terms
First term (a) = 3
Common difference (d) = 9/2 - 3 = 3/2
Sum of nth terms Sn, given n = 25
(iv) Given expression is, 41, 36, 31 to 12 terms.
First term (a) = 41
Common difference (d) = 36 - 41 = - 5
Sum of nth terms Sn, given n = 12
(v) a + b, a – b, a - 3b to 22 terms
First term (a) = a + b
Common difference (d) = a - b - a - b = -2b
Sum of nth terms Sn = n/2{2a(n - 1). d}
Here n = 22
S22 = 22/2{2.(a + b) + (22 - 1). -2b}
= 11{2(a + b) - 22b)
= 11{2a - 20b}
= 22a - 440b
∴S22 = 22a - 440b
(vi) (x - y)2,(x2 + y2), (x + y)2,... to n terms
First term (a) = (x - y)2
Common difference (d) = x2 + y2 - (x - y)2
= x2 + y2 - (x2 + y2 - 2xy)
= x2 + y2 - x2 + y2 + 2xy
= 2xy
= n/2{2(x - y)2 + (n - 1). 2xy}
= n{(x - y)2 + (n - 1)xy}
∴ Sn = n{(x — y)2 + (n — 1). xy)
(viii) Given expression -26, - 24. -22, to 36 terms
First term (a) = -26
Common difference (d) = -24 - (-26)
= -24 + 26 = 2
Sum of n tams Sn = n/2{2a + (n - 1)d)
Sum of n tams Sn = 36/2{2. -26 + (36 - 1)2}
= 18[-52 + 70]
= 18.18
= 324
∴ Sn = 324
Find the sum to n tam of the A.P. 5, 2, –1, – 4, –7, ...
Given AP is 5, 2, -1, -4, -7, .....
a = 5, d = 2 - 5 = -3
Sn = n/2{2a + (n - 1)d}
= n/2{2.5 + (n - 1) - 3}
= n/2{10 - 3(n - 1)}
= n/2{13 - 3n)
∴ Sn = n/2(13 - 3n)
Find the sum of n terms of an A.P. whose the terms is given by an = 5 - 6n.
Given nth term an = 5 - 6n
Put n = 1, a1 = 5 - 6.1 = -1
We know, first term (a1) = -1
Last term (an) = 5 - 6n = 1
Then Sn = n/2(-1 + 5 - 6n)
= n/2(4 - 6n) = n/2(2 - 3n)
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.
Given AP is 25, 22, 19, ..........
First term (a) = 25, d = 22 - 25 = -3.
Given, Sn = n/2(2a + (n - 1)d)
116 = n/2(2 × 25 + (n - 1) - 3)
232 = n(50 - 3(n - 1))
232 = n(53 - 3n)
232 = 53n - 3n2
3n2 - 53n + 232 = 0
(3n - 29)(n - 8) = 0
⟹ ag = 25 + (8 - 1) – 3
∴ n = 8, ag = 4
= 25 - 21 = 4
(i) How many terms of the sequence 18, 16, 14, ... should be taken so that their
(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?
(i) Given sequence, 18, 16, 14, ...
a = 18, d = 16 - 18 = -2.
Let, sum of n terms in the sequence is zero
Sn = 0
n/2(2a + (n. - 1)d) = 0
n/2(2.18 + (n - 1) - 2) = 0
n(18 - (n - 1)) = 0
n(19 - n) = 0
n = 0 or n = 19
(ii) ∵ n = 0 is not possible.
Therefore, sum of 19 numbers in the sequence is zero.
Given, a = -14, a 5 = 2
a+ (5 - 1)d = 2
-14 + 4d = 2
4d = 16 ⟹ d = 4
Sequence is -14, -10, -6, -2, 2, ...........
Given Sn = 40
40 = n/2{2(-14) + (n - 1)4}
80 = n(-28 + 4n - 4)
80 = n(-32 + 4n)
4(20) = 4n(-8 + n)
n2 - 8n - 20 = 0
(n - 10)(n + 2) = 0
n = 10 or n = -2
Sum of 10 numbers is 40 (Since -2 is not a natural number)
(iii) Given AP 9. 17, 25, ..................
a = 9, d = 17 - 9 = 8, and Sn = 636
636 = n/2(2.9 + (n - 1)8)
1272 = n(18 - 8 + 8n)
1272 = n(10 + 8n)
2 × 636 = 2n(5 + 4n)
636 = 5n + 4n2
4n2 + 5n - 636 = 0
(4n + 53)(n - 12) = 0
∴ n = 12 (Since n (-53)/4 is not a natural number)
Therefore, value of n is 12.
(iv) Given AP, 63, 60, 57, .............
a = 63, d = 60 - 63 = -3
Sn = 693
Sn = n/2(2a + (n - 1)d)
693 = n/2(2.63 + (n. - 1) - 3)
1386 = n(126 - 3n + 3)
1386 = (129 - 3n)n
3n2 - 129n + 1386 = 0
n2 - 43n + 462 = 0
n = 21, 22
∴ Sum of 71 or 77 tern is 693
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