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Question
If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
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Solution
Let a and d be the first term and common difference, respectively of an AP.
∵ Sum of n terms of an AP,
Sn = `n/2 [2a + (n - 1)d]` ...(i)
Now, S6 = 36 ...[Given]
⇒ `6/2[2a + (6 - 1)d]` = 36
⇒ 2a + 5d = 12 ...(ii)
And S16 = 256
⇒ `16/2[2a + (16 - 1)d]` = 256
⇒ 2a + 15d = 32 ...(iii)
On subtracting equation (ii) from equation (iii), we get
10d = 20
⇒ d = 2
From equation (ii),
2a + 5(2) = 12
⇒ 2a = 12 − 10 = 2
⇒ a = 1
∴ S10 = `10/2 [2a + (10 - 1)d]`
= 5[2(1) + 9(2)]
= 5(2 + 18)
= 5 × 20
= 100
Hence, the required sum of first 10 terms is 100.
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