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  • Given sum of the first n terms of an A.P. is . Another A.P. is formed with the same first term a

Given 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 of the common difference, the sum of n terms of the new A.P. is :

Option: 1

n+4n^{2}


Option: 2

6n^{2}-n


Option: 3

n^{2}+4n


Option: 4

3n+2n^{2}


Answers (1)

best_answer

S_n=2n+3n^{2}

T_n=S_n-S{n-1}=2n+3n^{2}-2(n-1)+3(n-1)^{2}

T_n=6n-1

Old Series 5,11,17,23.....

New Series 5,17,29.....

Sum of the new series

\frac{n}{2}[10+(n-1)12]=6n^2-n

 

Posted by

Divya Prakash Singh

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