Given the sum of the first n terms of an A.P.

Question

Given the sum of the first $n$ terms of an A.P. is $2n+3n^2$. Another A.P. is formed with the same first term and double the common difference; the sum of $n$ terms of the new A.P. is:

Accepted Answer

$a_1=S_1=5$; $a_2=S_2-S_1=16-5=11$, so original $d=6$. First term is unchanged ($a_1=5$); new common difference $d'=12$.

$$S_n'=\frac n2\big[2(5)+(n-1)(12)\big]=\frac n2(10+12n-12)=n(6n-1)=6n^2-n.$$

Solutions · 1