Question

Which term of the A.P.3,15,27,39,... will be 120 more than its 21st term?

Answer 1

Therefore the 31st term of the A.P will be 120 more than the its 21st term.

Step-by-step explanation:

Given arithmetic progression is

3,15,27,39,.....

The first term (a)=3

The common difference (d)= second term - first term

                                            =(15-3)

                                            =12

The nth term of an A.P series is

$T_n=a+(n-1)d$

The 21st term of the A.P is

$T_{21}=3+(21-1)\times12$

     =243

Let nth term of the A.P will be 120 more than its 21st term

$\therefore T_n=243+120$

$\Rightarrow a+(n-1)d= 363$

Putting the value of a and d

$\Rightarrow 3+(n-1)12=363$

$\Rightarrow (n-1)12=363-3$

$\Rightarrow (n-1) =\frac{360}{12}$

$\Rightarrow (n-1) =30$

$\Rightarrow n = 31$

Therefore the 31st term of the A.P will be 120 more than the its 21st term.