How to Find the Sum of Series in AP with Given Information ?
To find the sum of the series, we use the formula given below.
Sn = (n/2) [a + l] (or)
Sn = (n/2) [2a + (n - 1)d]
To find the first term, common difference and number of term, we use the formula given below.
an = a + (n - 1)d
Question 1 :
The first and last term of an AP are 17 and 350 respectively.If the common difference is 9, how many terms are there and what is their sum?
Solution :
a = 17 l = 350 d = 9
Sn = (n/2) [2a + (n-1)d]
Sn = (n/2) [2(17)+(n-1)9]
= (n/2) [34 + 9 n - 9]
= (n/2) [25 + 9 n] ----(1)
Sn = (n/2)[a + l]
= (n/2)[17 + 350] ------(2)
(1) = (2)
(n/2) [25 + 9 n] = (n/2)[17 + 350]
25 + 9 n = 17 + 350
9n = 367 - 25
9n = 342
n = 342/9
n = 38
38 terms are needed.
S38 = (38/2)[17 + 350]
= 19 [367]
= 6973
Question 2 :
The sum of first 22 terms of an AP in which d = 7and 22nd term is 149.
Solution :
n = 22 d = 7
a22 = a + 21 d
= a + 21 (7)
149 = a + 147
149 -147 = a
a = 2
Sn = (n/2) [2a + (n-1)d]
= (22/2) [2(2) + (22-1)7]
= 11 [4 + 21(7)]
= 11 [4 + 147]
= 11 [151]
= 1661
Question 3 :
The sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution :
a₂ = a + 1d
14 = a + d
a + d = 14 ---(1)
a₃ = a + 2d
18 = a + 2d
a + 2 d = 18 ------(2)
(1) - (2)
a + d = 14
a + 2 d = 18
(-) (-) (-)
--------------
- d = -4
d = 4
By applying the value of d = 4 in the (1), we get
a + 4 = 14
a = 14 - 4
a = 10
Sn = (n/2) [2a + (n - 1) d]
n = 51
Sn = (51/2)[2(10) + (51-1) 4]
= (51/2)[20 + 50(4)]
= (51/2) [20 + 200]
= (51/2) [220]
= 51 (110)
S51 = 5610
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