Question

In an AP:
(i) Given $a=5,d=3,a_n=50$, find $n$ and $S_n$.
(ii) Given $a=7,a_{13}=35$, find $d$ and $S_{13}$.
(iii) Given $a_{12}=37,d=3$, find $a$ and $S_{12}$.
(iv) Given $a_{13}=15,S_{10}=125$, find $d$ and $a_{10}$.

Answer 1

(i) Given $a=5,d=3,a_n=50$, find n and Sn .

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

$50=5+(n-1)3$

$15=n-1$

$n=16$

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

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

$=8\times 55$

$S_n=440$

(ii) Given $a=7,a_{13}=35$ , find d and $S_{13}$

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

$35=7+(13-1)d$

$28=12d$

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

$S_n=\frac{n}{2}[a+l]$

$=\frac{13}{2}[7+35]$

$=273$

$S_{13}=273$

(iii) Given $a_{12}=37,d=3$, find a and $S_{12}$

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

$37=a+(12-1)3$

$37-11\times 3=a$

$a=4$

$S_n=\frac{n}{2}[a+l]$

$=\frac{12}{2}[4+37]$

$=246$

$S_{12}=246$

(iv) Given $a_{13}=15,S_{10}=125$,find d and $a_{10}$

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

$15=a+(13-1)3$

$15=a+12d$...(1)

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

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

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

Solving (1) and (2), we get,

$d=-1,a=17$

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

$a_{10}=17+(10-1)(-1)$

$=17-9=8$

$a_{10}=8$