Question

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

• A. 13
• B. 12
• C. 5
• D. 6

Answer 1

The correct option is B 12

Solution : -

First A.P $25+29+33$

$n^{th}$ term $a_n=a+(n-1)d=25+(n-1)4$

$=25+4n-4$

$=4n+21$____A

Second A.P, $3+4+6+\mathrm{9....}$

Common deff is in A.P i.e, $1,2,3,...$

Now $n^{th}$ term $T_n=a{n}^2+bn+c$

$T_1=a+b+c=3$___(1)

$T_2=4a+2b+c=4$____(2)

$T_3=9a+3b+c=6$____(3)

Subtract (1) & (2), we get $3a+b=1$___(4)

By (2) & (3), we get $5a+b=2$___(5)

By (4) & (5), we get $2a=1$

$\therefore a=1/2$ $b=-1/2$ and $c=3$

Substituting in $T_n$ we get $T_n=\frac{n^2-n+6}{2}$___B

Equating A and B

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

$\Rightarrow n=12$