Sum to n terms of an Arithmetic progression is 2n2 + n then eighth term is
Let,last term(8th term)=l.
Eighth term be,substitute in question 2( $n^{2}$)+n then sum of eighth term=2( $8^{2}$)+8=136.
sum of 1st term, =2$(1^{2})$+1=3
a=3.
$s_{n}$=n/2[a+l]=136
8/2[a+l]=136
a+l=136/4=34
l=34-3=31
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