Given sum of the first $n$ terms of an A.P. is $2 n+$ $3 n^2$. Another A.P. is formed with the same first…
Given sum of the first $n$ terms of an A.P. is $2 n+$ $3 n^2$. Another A.P. is formed with the same first term and double of the common difference, the sum of $n$ terms of the new A.P. is :
$n+4 n^2$
$6 n^2-n$
$n^2+4 n$
$3 n+2 n^2$
Solution
Given $\mathrm{S}_n=2 n+3 n^2$
Now, first term $=2+3=5$
second term $=2(2)+3(4)=16$
third term $=2(3)+3(9)=33$
Now, sum given in option (b) only has the same first term and difference between 2 nd and 1 st term is double also.