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Question

In an AP:
(i) Given a=5,d=3, an=50, find n and Sn.
(ii) Given a=7, a13=35, find d and S13.
(iii) Given a12=37, d=3, find a and S12.
(iv) Given a3=15, S10=125, find d and a10.
(v) Given d=5, S9=75, find a and a9.
(vi) Given a=2,d=8, Sn=90, find n and an.
(vii) Given a=8, an=62, Sn=210, find n and d.
(viii) Given an=4, d=2, Sn=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.

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Solution

We know that in an AP

If first term is a1=a, common difference is d then the nth term is

an=a+(n1)d

Sum of first n terms of this AP is

Sn=n2(a1+an)=n2(2a+(n1)d)

Now,
(i)
a1=a=5
d=3
an=50=a+(n1)d
50=5+(n1)3
n=16

Sn=n2(a1+an)
Sn=162(5+50)
Sn=8×55=440

(ii)
a1=7
a13=35
n=13

a13=a1+(n1)d
35=7+(131)d
d=73

S13=132(a1+a13)
S13=132(7+35)
S13=13×21=273

(iii)
a12=37
d=3
n=12

a12=a1+(n1)d=a+(121)3
37=a+33
a=4

S12=n2(a1+a12)
s12=122(4+37)=6×41=256

(iv)
a3=15
a3=a+(31)d15=a+2d ...(1)

S10=125
102(2a+9d)=125
2a+9d=25 ....(2)

From (1) and (2), we get
d=1,a=17

a10=17+(9)(1)=8

(v)
d=5
S9=75=92(2a+8d)=92(2a+40)=9(a+20)

a=353

a9=a+8d=353+8×5=40353
a9=853

(vi)
a=2

d=8

Sn=90

n2(2a+(n1)d)=90

n2(8n4)=90

n(4n2)=90

2n2n45=0

n=5 (n>0, we have neglected negative value of n=92)

an=a5=a+(4)(d)=2+4×8=34

(vii)
a=8

an=62

Sn=210=n2(a+an)

420=n(8+62)=70n

n=6

an=a+(n1)d

62=8+(5)d

d=545

(viii)
an=4

d=2

an=a+(n1)d

4=a+(n1)2

=6=a+2n ...(1)

Sn=14

Sn=n2(a+an)

14=n2(a+4) ....(2)

From (1) and (2), we get

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

n=7, then a=8

(ix)
a=3

n=8

S=192

S=192=n2(2a+(n1)d)

48=6+7d

d=6

(x)
l=an=28

Sn=144

n=9

Sn=n2(a+an)

144=92(a+28)

a=4

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