Question

In an $A.P$ i) given $a=5,d=3,a_n=50$, find $n$ and $S_n$. ii) given $a_3=15,S_{10}=125$, find $d$ and $a_{10}$.

Answer 1

(i)

Given,

$a=5,d=3,a_n=50$

$a_n=a+(n-1)d\Rightarrow 50=5+(n-1)3$

$\Rightarrow 15=n-1\therefore n=16$

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

$=\frac{16}{2}\left[2\right(5)+(16-1\left)3\right]$

$=8[10+45]=440$

(ii)

Given,

$a_3=15,S_{10}=125$

$a+2d=15$.......(1)

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

$25=2a+9d$.......(2)

$\left(2\right)-2\times \left(1\right)$ gives,

$d=-1$

substituting the value of d in equation (1), we get,

$a=17$

$a_{10}=a+9d=17+9(-1)=8$