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The 14th term of an A.P. is twice its 8t...

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

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Let a be the first term and d be the common difference of the given AP. Then,
`T_(14) = 2 xx T_(8) rArr a +13d = 2(a +7d)`
`rArr a +d = 0 " "…(i)`
`"Also,"T_(6) = -8 rArr a +5d =-8. " "...(ii)`
On solving (i) and (ii), we get a = 2 and d = -2.
The sum of first 20 terms is given by
`S_(20) = (n)/(2) *[2a + (n-1)d,]"where" n = 20`
` = ((20)/(2)) xx {(2 xx 2 +19 xx (-2)}`
`= 10 xx (4-38) = 10 xx (-34) = -340.`
Hence, the required sum is -340.
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