Question

if the sum of the 3rd and 7th term of an arithmetic progression is 6 and their product is 8. Find the sum of first 20 terms of an arithmetic progression​

Answer 1

ANSWER

Let a and d be the first term and common difference of AP

nth term of AP

a

n

=a+(n−1)d

∴a

3

=a+(3−1)d=a+2d

a

7

=a+(7−1)d=a+6d

Given a

3

+a

7

=6

∴(a+2d)+(a+6d)=6

⇒2a+8d=6

⇒a+4d=3....(1)

Also given

a

3

×a

7

=8

∴(a+2d)(a+6d)=8

⇒(3−4d+2d)(3−4d+6d)=8 [Using (1)]

⇒(3−2d)(3+2d)=8

⇒9−4d

2

=8

⇒4d

2

=1

⇒d

2

=

4

1

⇒d=±

2

1

When d=

2

1

a=3−4d=3−4×

2

1

=3−2=1

When d=−

2

1

a=3−4d=3+4×

2

1

=3+2=5

When a=1 & d=

2

1

S

16

=

2

16

[2×1+(16−1)×

2

1

]=8(2+

2

15

)=4×19=76

When a=5 & d=−

2

1

S

16

=

2

16

[2×5+(16−1)×(−

2

1

)]=8(10−

2

15

)=4×5=20

Thus, the sum of first 16 terms of the AP is 76 or 20.