Question

If $n^{th}$ term of the series $25+29+33+37+....$ and $3+4+6+9+13+....$ are equal, the $n$ equals

• A. $11$
• B. $12$
• C. $13$
• D. $14$

Answer 1

The correct option is B $12$

Given series are

$25+29+33+37$ ……..(1)

$3+4+6+9+13$ ……….(2)

Now nth term of 1^st series is A.P.

Nth term is ,

$=25+(n-1)4$

$=21+4n$

Nth term of 2^nd series is $\frac{n^2-n+6}{2}$

$21+4n=\frac{n^2-n+6}{2}$

$42-8n=n^2-n+6$

$n^2+9n-36=0$

$n^2-12n+3n-36=0$

$n(n-12)+3(n-12)=0$

$(n-12)(n+3)=0$

$n=12,n=-3$

Taking $n=12$

Hence, this is the answer.