Question

Find the term of the arithmetic progression 9, 12, 15, 18, ... which is 39 more than its ${36}^{th}$ term.

• A. $55$th
• B. $40$th
  
• C. $19$th
• D. $49$th

Answer 1

The correct option is D $49$th

Given: $9,12,15,18,..$ are in A.P

where $a=9$ and $d=12-9=3$

$t_n=39+t_{36}$(given)

$t_{36}=a+35d=9+25\times 3=9+105=114$

$t_n=39+t_{36}=39+114=153$

$\Rightarrow a+(n-1)d=153$

$\Rightarrow 9+(n-1)\times 3=153$

$\Rightarrow (n-1)\times 3=153-9=144$

$\Rightarrow (n-1)=\frac{144}{3}=48$

$\Rightarrow n=48+1=49$

$\therefore$ the ${49}^{th}$ term is $153$ from above