Question

If the ${10}^{th}$ term and the ${18}^{th}$ term of an $A.P$ are $25$ and $41$ respectively then find the following $\left(a\right)$ the $1^{st}$ term and the common difference $\left(b\right)$ the ${38}^{th}$ term.

Answer 1

Given:- $a_{10}=25$

$a_{18}=41$

As we know that, $n^{th}$ term in an A.P. is given as

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

Here $a$ and $d$ are first term and common difference.

Therefore,

$a+9d=25.....\left(1\right)$

$a+17d=41.....\left(2\right)$

Subtracting $\left(1\right)$ from $\left(2\right)$, we have

$(a+17d)-(a+9d)=41-25$

$8d=16$

$\Rightarrow d=\frac{16}{8}=2$

Substituting the va;lue of $d$ in $\left(1\right)$, we have

$a+9\times 2=25$

$\Rightarrow a=25-18=7$

Hence first term and the common difference of A.P. are $7$ and $2$ respectively.

Also,

$a_{38}=a+(38-1)d$

$\Rightarrow a_{38}=7+37\times 2=81$

Hence ${38}^{th}$ term is $81$.