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 $.

AcademicMathematicsNCERTClass 10

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, $a=4$.

raja
Updated on 10-Oct-2022 13:20:30

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