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:

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$n + 4 n^{2}$

$3 n + 2 n^{2}$

$6 n^{2} - n$

$n^{2} + 4 n$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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

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