Question

An AP consists of $50$ terms of which $3rd$ term is $12$ and the last term is $106.$ Find the $29th$ term.

• A. $64$
  
• B. $72$
• C. $78$
• D. $80$

Answer 1

Answer: (A)

$64$

Let $a$ and $d$ be the first term and common difference of the given AP.
Given,$n=50$, last term $=106,a_3=12$

By using formula for nth term $t_n=a+(n-1)d$

$\Rightarrow a+2d=12$ ...(1)

$\Rightarrow a+49d=106$ ...(2)

On subtracting (1) from (2), we get

$47d=94\Rightarrow d=2$

Putting $d=2$ in (1), we get

$a=12-2\times 2=8$

Now, $a_{29}=a+28d=8+28\times 2=64$