Given sum of first n terms of an AP is 2n + 3n2. Another AP is formed with the same first term and double of the common difference, the sum of n terms of the new AP is:

A
$n + 4 n^{2}$
B
$6 n^{2} - n$
C
$n^{2} + 4 n$
D
$3 n + 2 n^{2}$

Solution:

$G i v e n s u m o f n t e r m s o f A P i s S_{n} = 2 n + 3 n^{2} F o r t h e f i r s t t e r m s u b s t i t u t e n = 1 H e n c e, a_{1} = 5 T h e s e c o n d t e r m i s a_{2} = S_{2} - S_{1} = 16 - 5 = 11 T h e c o m m o n d i f f e r e n c e i s a_{2} - a_{1} = 11 - 5 = 6 H e n c e, t h e f i r s t t e r m a n d c o m m o n d i f f e r e n c e o f t h e o r i g i n a l s e q u e n c e a r e a_{1} = 5, d = 6$

$\begin{array}{rcl} F o r t h e n e w s e q u n c e, f i r s t t e r m i s s a m e a & = & 5 o n l y \\ C o m m o n d i f f e r e n c e i s t w o t i m e s o f 6 \\ d & = & 12 \\ S u m o f f i r s t n t e r m s o f n e w A P i s S_{n} & = & \frac{n}{2} (2 a + (n - 1) d) \\ S_{n} & = & \frac{n}{2} (10 + (n - 1) 12) \\ = & 6 n (n) - n \\ = & 6 n^{2} - n \end{array}$