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Question
In an AP given a3 = 15, S10 = 125, find d and a10.
Solution
Given that, a3 = 15, S10 = 125
As an = a + (n − 1)d,
a3 = a + (3 − 1)d
15 = a + 2d ...(i)
Sn = `n/2 [2a + (n - 1)d]`
S10 = `10/2 [2a + (10 - 1)d]`
125 = 5(2a + 9d)
25 = 2a + 9d ...(ii)
On multiplying equation (i) by (ii), we get
30 = 2a + 4d ...(iii)
On subtracting equation (iii) from (ii), we get
−5 = 5d
d = −1
From equation (i),
15 = a + 2(−1)
15 = a − 2
a = 17
a10 = a + (10 − 1)d
a10 = 17 + (9) (−1)
a10 = 17 − 9
a10 = 8
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