Question

The ${10}^{th}$ and ${18}^{th}$ terms of an A.P. are $41$ and $73$ respectively. Find the ${27}^{th}$ term

Answer 1

It is given that the $10$th term of an A.P is $T_{10}=41$ and the $18$th term is $T_{18}=73$.

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$, therefore,

$T_{10}=a+(10-1)d\Rightarrow 41=a+9d.....\left(1\right)$

$T_{18}=a+(18-1)d\Rightarrow 73=a+17d.....\left(2\right)$

Now subtract equation (1) from equation (2) as follows:

$(a-a)+(17d-9d)=73-41$

$\Rightarrow 8d=32$

$\Rightarrow d=\frac{32}{8}$

$\Rightarrow d=4$

Substitute the value of $d$ in equation (1):

$a+(9\times 4)=41$

$\Rightarrow a+36=41$

$\Rightarrow a=41-36$

$\Rightarrow a=5$

Now, the ${27}^{th}$ term of an A.P with $a=5$ and $d=4$ is:

$T_{27}=5+(27-1)4=5+(26\times 4)=5+104=109$

Hence, the $27$th term of an A.P is $109$.