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