How to Find a and d of an Arithmetic Sequence :
To find the first term, common difference, number of terms and sum of an arithmetic series, we use one of the formulas given below.
an = a + (n - 1)d
Sn = (n/2) [a + l] (or)
Sn = (n/2) [2a + (n - 1)d]
Question 1 :
Given : a = 3 , n = 8 ,S n = 192 find d
Solution :
S n = (n/2) [2a + (n - 1) d]
192 = (8/2) [2 (3) + (8 - 1) d]
192 = 4 [6 + 7 d]
192/4 = [6 + 7 d]
48 - 6 = 7 d
42/7 = d
d = 6
Question 2 :
Given l = 28 , S = 144 , and there are total 9 terms. Find a
Solution :
S n = (n/2) [a + l]
n = 9
144 = (9/2) [ a + 28]
(144 x 2)/9 = a + 28
32 = a + 28
a = 32 - 28
a = 4
Question 2 :
How many terms of the AP 9, 17, 25,.......... must be taken to give a sum of 636?
Solution :
an = a + (n - 1) d
Sn = 636
Sn = (n/2) [2a + (n - 1) d]
a = 7 d = 17 - 9 = 8, Sn = 636
636 = (n/2)[2(9) + (n-1) 8]
(636 x 2) = n[18 + 8 n - 8]
(636 x 2) = n[10 + 8 n]
636 = n[5 + 4n]
636 = 5n + 4 n²
4 n² + 5 n - 636 = 0
(4n + 53)(n - 12) = 0
n = -53/4 n = 12
Hence 12 terms are need to give a sum of 636.
Question 3 :
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and common difference.
Solution :
a = 5, l = 45, Sn = 400
Sn = (n/2)[a + l]
400 = (n/2)[5 + 45]
400 = (n/2)[50]
400 = n(25)
400/25 = n
n = 16
Sn = (n/2)[2 a + (n- 1) d]
400 = (16/2) [2(5) + (16-1)d]
400 = 8 [10 + 15 d]
400/8 = 10 + 15 d
50 = 10 + 15 d
50 - 10 = 15 d
40 = 15 d
40/15 = d
d = 8/3
After having gone through the stuff given above, we hope that the students would have understood, how to find a and d of an arithmetic sequence given the sum.
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