Question

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

Answer 1

We know that the formula for the nth term is $t_n=a+(n-1)d$, where $a$ is the first term, $d$ is the common difference.

It is given that third term of an A.P is $t_3=12$, therefore,

$t_n=a+(n-1)d\Rightarrow 12=a+(3-1)d\Rightarrow a+2d=12......\left(1\right)$

Also, it is given that the $50$th term is $t_{50}=106$, therefore,

$t_n=a+(n-1)d\Rightarrow 106=a+(50-1)d\Rightarrow a+49d=106......\left(2\right)$

Now, subtract equation 1 from 2 as follows:

$(a-a)+(49d-2d)=106-12\Rightarrow 47d=94\Rightarrow d=\frac{94}{47}=2$

Substitute the value of the difference $d=2$ in equation 1:

$a+(2\times 2)=12\Rightarrow a+4=12\Rightarrow a=12-4=8$

Now, the $29$th term with $a=8$ and $d=2$ can be obtained as:

$t_{29}=8+(29-1)2=8+(28\times 2)=8+56=64$

Hence, the $29$th term is$64$.