In an AP :

In an AP :

(1) Given $a=5, d=3, a_{n}=50$, find $n$ and $S_{n}$.

(ii) Given $\mathrm{a}=7, \mathrm{a}_{13}=35$, find $\mathrm{d}$ and $\mathrm{S}_{13}$.

(iii) Given $\mathrm{a}_{12}=37, \mathrm{~d}=3$, find a and $\mathrm{S}_{12}$.

(iv) Given $a_{3}=15, S_{10}=125$, find $d$ and $a_{10}$.

(v) Given $\mathrm{d}=5, \mathrm{~S}_{9}=75$, find a and $\mathrm{a}_{9}$.

(vi) Given $\mathrm{a}=2, \mathrm{~d}=8, \mathrm{~S}_{\mathrm{n}}=90$, find $\mathrm{n}$ and $\mathrm{a}_{\mathrm{n}}$.

(vii) Given $\mathrm{a}=8, \mathrm{a}_{\mathrm{n}}=62, \mathrm{~S}_{\mathrm{n}}=210$, find $\mathrm{n}$ and $\mathrm{d}$.

(viii) Given $a_{n}=4, d=2, S_{n}=-14$, find $n$ and $a$.

(ix) Given $\mathrm{a}=3, \mathrm{n}=8, \mathrm{~S}=192$, find $\mathrm{d}$.

(x) Given $\ell=28, \mathrm{~S}=144$, and there are total 9 terms. Find $\mathrm{a}$.


(i) $a=5, d=3, a_{n}=50$

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

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

$\Rightarrow 5+3 n-3=50$ or $3 n=48$ or $n=16$

$\mathrm{S}_{16}=\frac{16}{2}\{2 \mathrm{a}+15 \mathrm{~d}\}$

$=8\{10+15 \times 3\}=440$

(ii) $a=7, a_{13}=35$

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

35 = 7 + 12 d

35 – 7 = 12d

28 = 12d




$=\frac{13 \times 42}{2}=13 \times 21$


(iii) $\mathrm{a}_{12}=37, \mathrm{~d}=3$

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


a = 4





(iv) $a_{3}=15, S_{10}=125$

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

$15=a+2 d$ …(i)

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$

$\mathrm{S}_{10}=\frac{10}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$

$125=5(2 a+9 d)$

$25=2 a+9 d$ ….(ii)

On multiplying equation (i) by 2 , we obtain

$30=2 a+4 d$ …(iii)

On subtracting equation (iii) from (ii), we obtain

$-5=5 \mathrm{~d}$


From equation (i),

15 = a + 2(–1)

15 = a – 2

a = 17

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



(v) $\mathrm{d}=5, \mathrm{~S}_{9}=75$

$\mathrm{S}_{9}=\frac{9}{2}[2 \mathrm{a}+(9-1) 5]$

$75=\frac{9}{2}(2 a+40)$

25 = 3(a + 20)

25 = 3a + 60

3a = 25 – 60






(vi) $\mathrm{a}=2, \mathrm{~d}=8, \mathrm{~S}_{\mathrm{n}}=90$ $\Rightarrow \frac{n}{2}\{2 a+(n-1) d\}=90$

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

$\Rightarrow \frac{n}{2}\{4+(n-1) \times 8\}=90$

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

$\Rightarrow 4 n^{2}-2 n-90=0$

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

$\Rightarrow 2 n^{2}-10 n+9 n-45=0$

$\Rightarrow(n-5)(2 n+9)=0$

$\Rightarrow \mathrm{n}-5=0 \quad(\because 2 \mathrm{n}+9 \neq 0)$

$\Rightarrow \mathrm{n}=5$

$a_{n}=a_{5}=a+4 d=2+4 \times 8=34$

$\Rightarrow a_{n}=34$

(vii) $a=8, a_{n}=62, S_{n}=210$




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

62 = 8 + (6 – 1)d

62 – 8 = 5d

54 = 5d


(viii) $a_{n}=4, d=2, S_{n}=-14$

Now, $a_{n}=4 \Rightarrow a+(n-1) d=4$

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

$\Rightarrow \mathrm{a}=6-2 \mathrm{n}$ ….(i)


$\Rightarrow \frac{n}{2}\{2 a+(n-1) d\}=-14$

$\Rightarrow \frac{n}{2}\{2(6-2 n)+(n-1)(2)\}=-14 \quad\{B y(1)\}$

$\Rightarrow \frac{n}{2}\{12-4 n+2 n-2)=-14$

$\Rightarrow \frac{n}{2}\{10-2 n\}=-14$

$\Rightarrow \mathrm{n}(\mathrm{n}-5)=14$

$\Rightarrow n^{2}-5 n-14=0$

$\Rightarrow \mathrm{n}^{2}-7 \mathrm{n}+2 \mathrm{n}-14=0$

$\Rightarrow \mathrm{n}(\mathrm{n}-7)+2(\mathrm{n}-7)=0$


$\Rightarrow \mathrm{n}=7$

From (1), $\mathrm{a}=6-2 \times 7=-8$


(ix) $\mathrm{a}=3, \mathrm{n}=8, \mathrm{~S}=192$

$192=\frac{8}{2}[2 \times 3+(8-1) \mathrm{d}]$

$192=4[6+7 \mathrm{~d}]$

$48=6+7 d$

$42=7 d$


(x) $\ell=28$, i.e., $\mathrm{t}_{\mathrm{n}}=28$

$\Rightarrow \mathrm{t}_{9}=28 \Rightarrow \mathrm{a}+8 \mathrm{~d}=28$

$\mathrm{S}=144$, i.e., $\mathrm{S}_{9}=144$

$\Rightarrow \frac{9}{2}\left\{\mathrm{t}_{1}+\mathrm{t}_{9}\right\}=144 \Rightarrow \frac{9}{2}(\mathrm{a}+28)=144$

$\Rightarrow a+28=32 \quad \Rightarrow a=4$

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