Question

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

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Answer 1

An AP consists of 50 terms and the ${50}^{th}$ term is equal to 106.

It is also given that $a_3=12$
Using formula $a_n=a+(n-1)d$ to find $n^{th}$ term of AP, we get
$a_{50}=a+(50-1)d$ and
$a_3=a+(3-1)d$
$\Rightarrow 106=a+(50-1)d$
$\Rightarrow 106=a+49d$
$\Rightarrow 12=a+(3-1)d$
$\Rightarrow 12=a+2d$
These are equations consisting of two variables. Let’s solve them using substitution method.
Using equation $106=a+49d$
We get $a=106-49d$
Putting value of a in the equation $12=a+2d$ we get
$12=106-49d+2d$
$\Rightarrow 47d=94$
$\Rightarrow d=\frac{94}{47}=2$
Putting value of d in the equation, $a=106-49d$ we get
$a=106-49\left(2\right)=106-98=8$
Therefore, First term, $a=8$ and Common difference, $d=2$
To find ${29}^{th}$ term we use formula $a_n=a+(n-1)d$ which is used to find $n^{th}$ term of AP, we get
$a_{29}=8+(29-1)2=8+56=64$
Therefore, ${29}^{th}$ term of AP is equal to 64.