Question

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

Answer 1

Given, $a_3=12$, $a_{50}=106$

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

$a_3=a+(3-1)d$

$12=a+2d$ ... (i)

$a_{50}=a+(50-1)d$

$106=a+49d$ ... (ii)

On subtracting (i) from (ii), we get

$94=47d$

$\therefore d=2$

From equation (i), we get

$12=a+2\left(2\right)$

$\Rightarrow a=12-4=8$

Now $a_{29}=a+(29-1)d$

$=8+\left(28\right)2$

$=8+56$

$=64$

Therefore, ${29}^{th}$ term is $64.$