﻿ The 14th term of an AP is twice its 8th term. If its 6th  terms

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# 11.The 14th term of an AP is twice its 8th term. If its 6th  terms is -8, then find the sum of its first 20 terms.

Let a and d be the first term and the common difference of the AP, respectively.

∴ nth term of the AP, an = a + (n-1)d

So,

a14 = a+ (14-1)d = a + 13d

a8 = a + (8 - 1)d = a + 7d

a6 = a+ (6 - 1)d = a + 5d

According to the question,

a14 = 2a8

⇒ a + 13d = 2 (a + 7d)

⇒ a + d = 0 .... (i)

Also,

a6 = a + 5d = - 8 ... (ii)

solving (i) and (ii), we get

a = 2 and d = -2

∴ S20 = 20/2 [2 x 2 + (20 - 1)(-2)]

= - 340

Hence, the sum of the first 20 terms is -340.

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12.

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