Let `x_(1),x_(2),…..,x_(n)` be is an AP . If `x_(1)+x_(4)+x_(9)+x_(11)+x_(20)+x_(22)+x_(27)+x_(30)= 272 ,` then `x_(1)+x_(2)+x_(3)+…+x_(30)` is equal to

Options

A. 1020
B. 1200
C. 716
D. 2720

Answer

a

Explanation

Given , `x_(1),x_(2), ….x_(n)` are A P .
We know that in an AP , the sum of the terms equidistant from the beginning and end ia always same and is equal to the sum of first and last term i.e
`a_(1) +a_(n) = a_(2) +a_(n-1)=a_(3) +a_(n-2)= ...`
If an AP consists of 30 terms , then
`x_(1) +x_(30)=x_(4)+x_(27)=x_(9)+x_(22) = x_(11) = x_(11) +x_(20)`
` :' x_(1) +x_(4) +x_(9) +x_(11) +x_(20) +x_(22)+x_(27)+x_(30)= 272 ` [ given]
` rArr (x_(1)+x_(30))+(x_(4)+x_(27)) +(x_(9)+x_(20))=272`
`rArr 4 (x_(1)+x_(30))= 272 rArr x_(1) +x_(30)= 272/4 = 68`
`S_(30)=30/2 (x_(1)+x_(30)) = 15(68) = 1020`