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If `a_(1), a_(2), a_(3)`,... are in AP such that `a_(1) + a_(7) + a_(16) = 40`, then the sum of the first 15 terms of this AP is

A

200

B

280

C

120

D

150

Text Solution

Verified by Experts

The correct Answer is:
A
Let the common difference of given AP is 'd'.
Since, `a_(1) + a_(7) + a_(16) = 40`
`:. a_(1) + a_(1) + 6d + a_(1) + 15d = 40 " " [ :' a_(n) = a_(1) + (n-1)d]`
`rArr 3a_(1) + 21d = 40`...(i)
Now, sum of first 15 terms is given by
`S_(16) = (15)/(2) [2a_(1) + (15 -1) d]`
`= (15)/(2) [2a_(1) + 14d] = 15 [a_(1) + 7d]`
From Eq. (i), we have
`a_(1) + 7d = (40)/(3)`
So, `S_(16) = 15 xx (40)/(3)`
`= 5 xx 40 = 200`
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IIT JEE PREVIOUS YEAR-SEQUENCES AND SERIES-SUM OF N TERMS OF AN AP
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