Question

Given , $a_3=15,s_{10}=125$ find $d$ and $a_{10}$.

Answer 1

Sincem $n^{th}$ term in A.P.,

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

$\Rightarrow a_3=a+2d$

$\Rightarrow a+2d=15$

$\Rightarrow a=15-2d$---(1)

$\Rightarrow s_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$\Rightarrow 125=\frac{10}{2}[2a+(10-1\left)d\right]$

$\Rightarrow 125\times \frac{2}{10}=[2a+(10-1\left)d\right]$

$\Rightarrow 2a+9d=25$

$\Rightarrow 2(15-2d)+9d=25$ [Since from (1)]

$\Rightarrow 30-4d+9d=25$

$\Rightarrow 5d=-5$

$\Rightarrow d=-1$

$\Rightarrow a=15-2d=15-2(-1)$

$\Rightarrow a=17$

Now, the $a_{10}=a+9d$

$=17+9(-1)=8$

Hence, $a_{10}=8$

Therefore, $d=-1$ and $a_{10}=8$.