Question

The sum of the third and seventh term of an A.P. is 8. Then
the sum of the first nine terms of this progression is

Answer 1

Step-by-step explanation:

an =a+(n−1)d

∴a 3 =a+(3−1)d=a+2d

a7 =a+(7−1)d=a+6d

Given a 3 +a7 =6

∴(a+2d)+(a+6d)=6

⇒2a+8d=6

⇒a+4d=3....(1)

Also givena a3 ×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= 1 / 4

⇒d=± 1 /2

When d= 1 /2

a=3−4d=3−4× 1/2 =3−2=1

When d=− 1 /2

a=3−4d=3+4× -1 /2 =3+2=5

When a=1 & d= 1/2

S 16 = 16 / 2 [2×1+(16−1)× 1 /2 ]=8(2+ 15 / 2 )=4×19=76

When a=5 & d=−1 /2

S 16 = 16 / 2 [2×5+(16−1)×(− 1 /2 )]=8(10− 15/2 )=4×5=20

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

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