Question

In an A.P
i) Given $a=2,d=8,Sn=90$, Find n & $a_9$
ii) Given $a=7,a_{1}3=35$, Find d & $S_{13}$.

Answer 1

$i)Givena=2,d=8,S_n=90$

Sum of n terms of an A. P is given by:

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

$S_n=na+\frac{n}{2}(n-1)d$

$90=2n+\frac{n}{2}(n-1)8$

$4n(n-1)+2n=90$

$4n^2-4n+2n=90$

$4n^2-2n-90=0$

solving above quadratic we get:

$n=\frac{2\pm \sqrt{4+1440}}{8}$

$n=\frac{2\pm \sqrt{1444}}{8}=\frac{2\pm 38}{8}$

$n=\frac{40}{8}=5$ or $n=\frac{-36}{8}$

Since, number of terms (n) can not be negative.

Hence, $n=5$

The nth term of an A.P is given by:

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

$a_9=2+(9-1)8$

$a_9=2+8\times 8$

$a_9=2+64$

Hence,

$a_9=66$

$ii)Givena=7,a_{13}=35$

The nth term of an A.P is given by:

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

$a_{13}=a+(13-1)d$

$35=7+12d$

$d=\frac{35-7}{12}=\frac{28}{12}$

Hence,

$d=\frac{7}{3}$

Sum of n terms of an A. P is given by:

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

$S_{13}=\frac{13}{2}(2\times 7+(13-1)\times \frac{7}{3})$

$S_{13}=\frac{13}{2}(14+12\times \frac{7}{3})$

$S_{13}=\frac{13}{2}(14+28)$

$S_{13}=\frac{13}{2}\left(42\right)$

Hence,

$S_{13}=273$