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 $a_3=15$, $S_{10}=125$, find $d$ and $a_{10}$.
(v) Given $d=5$, $S_9=75$, find $a$ and $a_9$.
(vi) Given $a=2,d=8$, $S_n=90$, find $n$ and $a_n$.
(vii) Given $a=8$, $a_n=62$, $S_n=210$, find $n$ and $d$.
(viii) Given $a_n=4$, $d=2$, $S_n=-14$, find $n$ and $a$.
(ix) Given $a=3,n=8,S=192$, find $d$.
(x) Given $l=28,S=144$, and there are total 9 terms. Find $a$.

Answer 1

We know that in an AP

If first term is $a_1=a$, common difference is $d$ then the $n^{th}$ term is

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

Sum of first $n$ terms of this AP is

$S_n=\frac{n}{2}(a_1+a_n)=\frac{n}{2}(2a+(n-1\left)d\right)$

Now,

(i)

$a_1=a=5$

$d=3$

$a_n=50=a+(n-1)d$

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

$\Rightarrow n=16$

$S_n=\frac{n}{2}(a_1+a_n)$

$\Rightarrow S_n=\frac{16}{2}(5+50)$

$\Rightarrow S_n=8\times 55=440$

(ii)

$a_1=7$

$a_{13}=35$

$\Rightarrow n=13$

$a_{13}=a_1+(n-1)d$

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

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

$S_{13}=\frac{13}{2}(a_1+a_{13})$

$\Rightarrow S_{13}=\frac{13}{2}(7+35)$

$\Rightarrow S_{13}=13\times 21=273$

(iii)

$a_{12}=37$

$d=3$

$\Rightarrow n=12$

$a_{12}=a_1+(n-1)d=a+(12-1)3$

$\Rightarrow 37=a+33$

$\Rightarrow a=4$

$S_{12}=\frac{n}{2}(a_1+a_{12})$

$\Rightarrow s_{12}=\frac{12}{2}(4+37)=6\times 41=256$

(iv)

$a_3=15$

$\Rightarrow a_3=a+(3-1)d\Rightarrow 15=a+2d$ ...(1)

$S_{10}=125$

$\frac{10}{2}(2a+9d)=125$

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

From (1) and (2), we get

$d=-1,a=17$

$a_{10}=17+\left(9\right)(-1)=8$

(v)

$d=5$

$S_9=75=\frac{9}{2}(2a+8d)=\frac{9}{2}(2a+40)=9(a+20)$

$\Rightarrow a=-\frac{35}{3}$

$a_9=a+8d=-\frac{35}{3}+8\times 5=40-\frac{35}{3}$

$a_9=\frac{85}{3}$

(vi)

$a=2$

$d=8$

$S_n=90$

$\Rightarrow \frac{n}{2}(2a+(n-1\left)d\right)=90$

$\Rightarrow \frac{n}{2}(8n-4)=90$

$\Rightarrow n(4n-2)=90$

$\Rightarrow 2n^2-n-45=0$

$\Rightarrow n=5$ $(\because n>0,\therefore$ we have neglected negative value of $n=-\frac{9}{2})$

$a_n=a_5=a+\left(4\right)\left(d\right)=2+4\times 8=34$

(vii)

$a=8$

$a_n=62$

$S_n=210=\frac{n}{2}(a+a_n)$

$\Rightarrow 420=n(8+62)=70n$

$\Rightarrow n=6$

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

$62=8+\left(5\right)d$

$\Rightarrow d=\frac{54}{5}$

(viii)

$a_n=4$

$d=2$

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

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

$\Rightarrow =6=a+2n$ ...(1)

$S_n=-14$

$S_n=\frac{n}{2}(a+a_n)$

$-14=\frac{n}{2}(a+4)$ ....(2)

From (1) and (2), we get

$n=7$ or $n=-2$, ($\because n>0,$ we have to neglect negative value)

$n=7,$ then $a=-8$

(ix)

$a=3$

$n=8$

$S=192$

$S=192=\frac{n}{2}(2a+(n-1\left)d\right)$

$\Rightarrow 48=6+7d$

$\Rightarrow d=6$

(x)

$l=a_n=28$

$S_n=144$

$n=9$

$S_n=\frac{n}{2}(a+a_n)$

$144=\frac{9}{2}(a+28)$

$\Rightarrow a=4$