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In an AP:
(i) given $ a=5, d=3, a_{n}=50 $, find $ n $ and $ S_{n^{\circ}} $.
(ii) given $ a=7, a_{13}=35 $, find $ d $ and $ \mathrm{S}_{13} $.
(iii) given $ a_{12}=37, d=3 $, find $ a $ and $ \mathrm{S}_{12} $.
(iv) given $ a_{3}=15, \mathrm{~S}_{10}=125 $, find $ d $ and $ a_{10} $
(v) given $ d=5, \mathrm{~S}_{9}=75 $, find $ a $ and $ a_{9} $.
(vi) given $ a=2, d=8, \mathrm{~S}_{n}=90 $, find $ n $ and $ a_{n} $
(vii) given $ a=8, a_{n}=62, \mathrm{~S}_{\mathrm{n}}=210 $, find $ n $ and $ d $.
(viii) given $ a_{n}=4, d=2, \mathrm{~S}_{n}=-14 $, find $ n $ and $ a $.
(ix) given $ a=3, n=8, \mathrm{~S}=192 $, find $ d $.
(x) given $ l=28, S=144 $, and there are total 9 terms. Find $ a $.
To do:
We have to find the given values in each case.
Solution:
(i) We know that,
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$n$th term $a_n=a+(n-1)d$
This implies,
$a_n=5+(n-1)3$
$50=5+(n-1)3$
$50-5=(n-1)3$
$45=(n-1)3$
$n-1=15$
$n=15+1$
$n=16$
$S_n=\frac{16}{2}[2 \times 5+(16-1) \times 3]$
$=8[10+15 \times 3]$
$=8(10+45)$
$=8 \times 55$
$=440$
Therefore, $n=16$ and $S_n=440$.
(ii) We know that,
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$a_{13}=35$
$a+12 d=35$
$7+12 d=35$
$12d=35-7$
$d=\frac{28}{12}$
$d=\frac{7}{3}$
$S_{13}=\frac{13}{2}[a+a_{13}]$
$=\frac{13}{2}[7+35]$
$=\frac{13}{2}(42)$
$=13 \times 21$
$=273$
(iii) We know that,
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$a_{12}=37$
$a+11 d=37$
$a+11(3)=37$
$a=37-33$
$a=4$
Therefore,
$S_{12}=\frac{12}{2}[a+a_{12}]$
$=6[4+37]$
$=6 \times 41$
$=246$
(iv) We know that,
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$a_{3}=15$
$a+2d=15$
$a=15-2d$......(i)
$S_{10}=\frac{10}{2}[2a+(10-1)d]$
$=5[2(15-2d)+9d]$
$=5[30-4d+9d]$
$=5(30+5 d)$
$125=5(30+5d)$
$\frac{125}{5}=30+5d$
$25-30=5d$
$-5=5 d$
$d=-1$
This implies,
$a=15-2(-1)$
$=15+2$
$=17$
$a_{10}=a+(10-1)(-1)$
$=17+9(-1)$
$=17-9$
$=8$
(v) We know that,
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$S_{9}=\frac{9}{2}[2 a+(9-1) d]$
$75=\frac{9}{2}[2 a+8 d]$
$75 \times \frac{2}{9}=2a+8(5)$
$\frac{50}{3}=2 a+40$
$2a=\frac{50}{3}-40$
$2a=\frac{50-120}{3}$
$2a=\frac{-70}{3}$
$a=\frac{-35}{3}$
$a_{9}=a+8 d$
$=\frac{-35}{3}+8(5)$
$=\frac{-35}{3}+40$
$=\frac{-35+120}{3}$
$=\frac{85}{3}$
(vi) We know that,
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$n$th term $a_n=a+(n-1)d$
This implies,
$a_n=2+(n-1)8$
$=2+8n-8$
$=8n-6$........(i)
$S_n=\frac{n}{2}[2 \times 2+(n-1)8]$
$90=\frac{n}{2}[4+8n-8]$ (From (i))
$90(2)=n(8n-4)$
$180=4n(2n-1)$
$n(2n-1)=45$
$2n^2-n-45=0$
$2n^2-10n+9n-45=0$
$2n(n-5)+9(n-5)=0$
$(2n+9)(n-5)=0$
$n=5$ or $2n=-9$ which is not possible as $n$ cannot be negative
$\therefore n=5$
This implies,
$a_n=8(5)-6$
$=40-6$
$=34$
Therefore, $n=5$ and $a_n=34$.
(vii) We know that,
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$n$th term $a_n=a+(n-1)d$
This implies,
$a_n=8+(n-1)d$
$62=8+(n-1)d$
$62-8=(n-1)d$
$54=(n-1)d$
$(n-1)d=54$........(i)
$S_n=\frac{n}{2}[2 \times 8+(n-1)d]$
$210=\frac{n}{2}[16+54]$ (From (i))
$210(2)=n(70)$
$3(2)=n$
$n=6$
$\therefore (6-1)d=54$
$5d=54$
$d=\frac{54}{5}$
Therefore, $n=6$ and $d=\frac{54}{5}$.
(viii) We know that,
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$n$th term $a_n=a+(n-1)d$
This implies,
$a_n=a+(n-1)2$
$4=a+(n-1)2$
$4-2n+2=a$
$a=6-2n$......(i)
$S_n=\frac{n}{2}[2 \times a+(n-1) \times 2]$
$-14=\frac{n}{2} \times 2[a+(n-1)]$
$-14=n(6-2n+n-1)$
$-14=n(5-n)$
$-14=5n-n^2$
$n^2-5n-14=0$
$n^2-7n+2n-14=0$
$n(n-7)+2(n-7)=0$
$(n-7)(n+2)=0$
$n=7$ or $n=-2$ which is not possible as $n$ cannot be negative
$\therefore a=6-2(7)$
$=6-14$
$=-8$
Therefore, $a=-8$ and $n=7$.
(ix) We know that,
$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$
$S_n=\frac{8}{2}[2 \times 3+(8-1) \times d]$
$192=4[6+7d]$
$48=(6+7d)$
$7d=48-6$
$7d=42$
$d=\frac{42}{7}$
$d=6$
Therefore, $d=6$.
(x) $a_n=l=28$
We know that,
$\mathrm{S}_{n}=\frac{n}{2}[a+l]$
$S_n=\frac{9}{2}[a+28]$
$144(2)=9(a+28)$
$16(2)=a+28$
$a=32-28$
$a=4$
Therefore, $a=4$.
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