Question

In an AP, it is given that $T_8=31$ and $T_{15}=45.$ Find the AP.

Answer 1

Given: $T_8=31$ and $T_{15}=45$

Let the first term of the given AP be $a$ and the common difference be $d.$

We know that $n^{th}$ term of an AP is given by $T_n=a+(n-1)d$
Then we have:
$T_8=a+7d=31\dots \dots \left(i\right)$
$T_{15}=a+14d=45\dots \dots \left(ii\right)$
Subtracting $eq.\left(i\right)$ from $\left(ii\right),$ we get

$a+14d-(a+7d)=45-31$

$\Rightarrow 14d-7d=14$
$\Rightarrow 7d=14$
$\Rightarrow d=2$
On putting the value of $d$ in $eq.\left(i\right),$ we get
$a+7\times 2=31$

$\Rightarrow a=31-14=17$

so, the terms of the given AP will be

$17,(17+2=19),(19+2=21),.......$
$\therefore$ The given AP is $17,19,21,23,.....$