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Question

In an AP
(i) given a=5,d=3,an=50, find n and Sn.
(ii) given a12=37,d=3, find a and S12.
(iii) given d=5,S9=75, find a and a9.
(iv) given a=8,an=62,Sn=210, find n and d.
(v) given l=28,S=144, and there are total of 9 terms. Find a.


Solution

(i) Given a=5, d=3, an=50, find n and Sn.

Using formula an=a+ (n−1) d,   to find  nth term of arithmetic progression, we can say that

an=5+ (n−1)(3)

50=5+3n−3

48=3n

n=483=16

Applying formula, Sn=n2(2a+ (n−1)d) to find sum of n terms of AP, we get

S16=16/2(10+ (16−1) (3)) =8(10+45) =8(55) =440

Therefore, n=16 and Sn=440

(ii) Given a12=37, d=3, find a and S12.

Using formula an=a+ (n−1) d,   to find nth term of arithmetic progression, we can say that

a12=a + (12−1) 3

37=a+33

a=4

Applying formula, Sn=n2(2a + (n−1) d) to find sum of n terms of AP, we get

S12=122(8+ (12−1) (3) =6(8+33) =6(41) =246

Therefore, a=4 and S12=246

(iii) Given d=5, S9=75, find a and a9.

Applying formula, Sn=n2 (2a+ (n−1) d) to find sum of n terms of AP, we get

S9=92 (2a+ (9−1) 5)

75=92 (2a+40)

150=18a+360

−210=18a

a=−21018 =−353 

Using formula an=a+ (n−1) d,   to find nth term of arithmetic progression, we can say that

a9=−353 + (9−1) (5) =−353 +40 = (35+120)3 = 853 

Therefore, a=353 and a9=853

(iv) Given a=8, an=62, Sn=210, find n and d.

Using formula an=a+ (n−1) d,   to find nth term of arithmetic progression, we can say that

62=8+ (n−1) (d) =8+nd−d

62=8+nd−d

nd−d=54

nd=54+d                    … (1)

Applying formula, Sn=n2 (2a+ (n−1) d) to find sum of n terms of AP, we get

210=n2 (16+ (n−1) d) =n2 (16+nd−d)

Putting equation (1) in the above equation, we get

210=n2 (16+54+d−d) =n2 (70)

21070 =n2 

n=6

Putting value of n in equation (1), we get

6d=54+d

d=545 

Therefore, n=6 and d=545 

(v) Given l=28, S=144, and there are total of 9 terms. Find a.

Applying formula, Sn=n2 [a +l], to find sum of n terms, we get

144=92 [a+28]

2889 =[a+28]

32=a+28

a=4


Mathematics

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