Question

The sum of the third and the seventh terms of an AP is 6 and their products is 8. Find the sum of the first sixteen terms of the AP.​

Answer 1

given a+2d+a+6d=6

=>2a +8d=6

=>2a=6-8d

=>a=3-4d

and given

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

putting the vale of a we get

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

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

=>9-4d^2=8

=>-4d^2=-1

=>d=1/2

so.a=3-1/2

=5/2

so sum of the 1st 16 turm =

16/2[2x1/2+(16-1)5/2]

=8[1+75/2]

=8[77/2]

=308

Answer 2

$<marquee direction="up" >⬆[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$ ⬆</marquee >[/tex]

given a+2d+a+6d=6

=>2a +8d=6

=>2a=6-8d

=>a=3-4d

and given

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

putting the vale of a we get

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

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

=>9-4d^2=8

=>-4d^2=-1

=>d=1/2

so.a=3-1/2

=5/2

so sum of the 1st 16 turm =

16/2[2x1/2+(16-1)5/2]

=8[1+75/2]

=8[77/2]

=308