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Question 128

If the sum of first 21 terms of an arithmetic progression is 609, then the $$11^{th}$$ term of progression is

Solution

Sn=n/2[2a+(n-1)d]

609=21/2[2a+(21-1)d]

by taking 2 common and eliminate 2

609/21=a+10d i.e a11

a11=29

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