Question.
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 $\mathrm{a}$.
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 $\mathrm{a}$.
Solution:
(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
$\mathrm{d}=\frac{7}{3}$
$S_{13}=\frac{n}{2}\left[a+a_{13}\right]$
$=\frac{13}{2}[7+35]$
$=\frac{13 \times 42}{2}=13 \times 21$
$=273$
(iii) $\mathrm{a}_{12}=37, \mathrm{~d}=3$
$a_{12}=a+(12-1) 3$
$37=a+33$
a = 4
$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$
$\mathrm{S}_{12}=\frac{12}{2}[4+37]$
$\mathrm{S}_{12}=6(41)$
$S_{12}=246$
(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
$-5=5 \mathrm{~d}$
$d=-1$
From equation (i),
15 = a + 2(–1)
15 = a – 2
a = 17
$a_{10}=a+(10-1) d$
$a_{10}=17+(9)(-1)$
$a_{10}=17-9=8$
(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
$a=\frac{-35}{3}$
$a_{9}=a+(9-1)(5)$
$=\frac{-35}{3}+8(5)$
$=\frac{-35}{3}+40$
$=\frac{-35+120}{3}=\frac{85}{3}$
(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) $a=8, a_{n}=62, S_{n}=210$
$210=\frac{n}{2}[8+62]$
$210=\frac{n}{2}(70)$
$n=6$
$a_{n}=a+(n-1) d$
62 = 8 + (6 – 1)d
62 – 8 = 5d
54 = 5d
$\mathrm{d}=\frac{54}{5}$
(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)
$S_{n}=-14$
$\Rightarrow \frac{n}{2}\{2 a+(n-1) d\}=-14$
$\Rightarrow \frac{n}{2}\{2(6-2 n)+(n-1)(2)\}=-14 \quad\{B y(1)\}$
$\Rightarrow \frac{n}{2}\{12-4 n+2 n-2)=-14$
$\Rightarrow \frac{n}{2}\{10-2 n\}=-14$
$\Rightarrow \mathrm{n}(\mathrm{n}-5)=14$
$\Rightarrow n^{2}-5 n-14=0$
$\Rightarrow \mathrm{n}^{2}-7 \mathrm{n}+2 \mathrm{n}-14=0$
$\Rightarrow \mathrm{n}(\mathrm{n}-7)+2(\mathrm{n}-7)=0$
$\Rightarrow(n-7)(n+2)=0$
$\Rightarrow \mathrm{n}=7$
From (1), $\mathrm{a}=6-2 \times 7=-8$
$a=-8$
(ix) $\mathrm{a}=3, \mathrm{n}=8, \mathrm{~S}=192$
$192=\frac{8}{2}[2 \times 3+(8-1) \mathrm{d}]$
$192=4[6+7 \mathrm{~d}]$
$48=6+7 d$
$42=7 d$
$d=6$
(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., $\mathrm{S}_{9}=144$
$\Rightarrow \frac{9}{2}\left\{\mathrm{t}_{1}+\mathrm{t}_{9}\right\}=144 \Rightarrow \frac{9}{2}(\mathrm{a}+28)=144$
$\Rightarrow a+28=32 \quad \Rightarrow a=4$
(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
$\mathrm{d}=\frac{7}{3}$
$S_{13}=\frac{n}{2}\left[a+a_{13}\right]$
$=\frac{13}{2}[7+35]$
$=\frac{13 \times 42}{2}=13 \times 21$
$=273$
(iii) $\mathrm{a}_{12}=37, \mathrm{~d}=3$
$a_{12}=a+(12-1) 3$
$37=a+33$
a = 4
$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$
$\mathrm{S}_{12}=\frac{12}{2}[4+37]$
$\mathrm{S}_{12}=6(41)$
$S_{12}=246$
(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
$-5=5 \mathrm{~d}$
$d=-1$
From equation (i),
15 = a + 2(–1)
15 = a – 2
a = 17
$a_{10}=a+(10-1) d$
$a_{10}=17+(9)(-1)$
$a_{10}=17-9=8$
(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
$a=\frac{-35}{3}$
$a_{9}=a+(9-1)(5)$
$=\frac{-35}{3}+8(5)$
$=\frac{-35}{3}+40$
$=\frac{-35+120}{3}=\frac{85}{3}$
(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) $a=8, a_{n}=62, S_{n}=210$
$210=\frac{n}{2}[8+62]$
$210=\frac{n}{2}(70)$
$n=6$
$a_{n}=a+(n-1) d$
62 = 8 + (6 – 1)d
62 – 8 = 5d
54 = 5d
$\mathrm{d}=\frac{54}{5}$
(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)
$S_{n}=-14$
$\Rightarrow \frac{n}{2}\{2 a+(n-1) d\}=-14$
$\Rightarrow \frac{n}{2}\{2(6-2 n)+(n-1)(2)\}=-14 \quad\{B y(1)\}$
$\Rightarrow \frac{n}{2}\{12-4 n+2 n-2)=-14$
$\Rightarrow \frac{n}{2}\{10-2 n\}=-14$
$\Rightarrow \mathrm{n}(\mathrm{n}-5)=14$
$\Rightarrow n^{2}-5 n-14=0$
$\Rightarrow \mathrm{n}^{2}-7 \mathrm{n}+2 \mathrm{n}-14=0$
$\Rightarrow \mathrm{n}(\mathrm{n}-7)+2(\mathrm{n}-7)=0$
$\Rightarrow(n-7)(n+2)=0$
$\Rightarrow \mathrm{n}=7$
From (1), $\mathrm{a}=6-2 \times 7=-8$
$a=-8$
(ix) $\mathrm{a}=3, \mathrm{n}=8, \mathrm{~S}=192$
$192=\frac{8}{2}[2 \times 3+(8-1) \mathrm{d}]$
$192=4[6+7 \mathrm{~d}]$
$48=6+7 d$
$42=7 d$
$d=6$
(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., $\mathrm{S}_{9}=144$
$\Rightarrow \frac{9}{2}\left\{\mathrm{t}_{1}+\mathrm{t}_{9}\right\}=144 \Rightarrow \frac{9}{2}(\mathrm{a}+28)=144$
$\Rightarrow a+28=32 \quad \Rightarrow a=4$