Question

0.5 The sum of the 3rd & 7th term of Ap
is 6 their products is 8. find the sum of
first 16th term of the A.P.​

Answer 1

Answer:

Given:

a(3)+a(7)=6

a+(3-1)d+a+(7-1)d=6

2a+2d+6d=6

2a+8d=6

2(a+4d)=6

a+4d=6/2

a+4d=3

a=3-4d__(1)

and also given that

(a+2d)(a+6d)=8

substitute'a' value in the equation

(3-4d+2d)(3-4d+6d)=8

(3-2d)(3+2d)=8

3²-(2d)²=8 (a+b)(a- b)=a²-b²

9-4d²=8

-4d²=8-9

d²=-1/-4

d=1/2

substitute d value in (1)

a=3-4(1/2)

a=1

s(16)=n/2[2a+(n-1)d]

=16/2[2(1)+(16-1)1/2]

=8[2+15/2]

=8(19/2)

=72

Therefore sum of first 16 terms=72