(Q2).The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Let the 3rd term of AP = a + 2d

and the 7th term of AP = a + 6d

∴ Sum of 3rd and 7th terms = a + 2d + a + 6d = 6

⇒ a + d =

⇒ a + d = …… (1)

Now product of above two terms = (a + 2d) (a + 6d) = 8

we can re-write above terms as following

(a + 4d – 2d) (a + 4d + 2d) = 8

⇒ ( – d) ( + d) =

⇒ – d2 =

⇒ d2 = – =

∴ d2 =
1
⇒ d = ±
1
 ....... (2)
Now putting d =
1
 in eq(1) we get
a + 4d = 3 ⇒ a + 4(
1
 ) = 3 ⇒ a =
so a = , d =
1
(or) now putting d =
-
 , we get
a + 4d = 3 ⇒ a + 4(
-
 ) = 3

⇒ a – =

⇒ a =

∴ a = , d =
-

∴ Sum of sixteen terms =

S16 =
2
 ( 2() + (16-)(
1
 ) )
= [ +
2
 ] =
×
2
 =
(or) = S16 =
2
 [ 2() + (16-)(
-1
 ) ]
= [ -
2
 ]
=
(+)
2
 = +

So S16 = or