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)
aya+(7-1)da+6d
given a_{3} + a_{7} = 6
* (a + 2d) + (a + 6d) = 6
rightarrow 2a + 8 * d = 6
rightarrow a+4d=3.... (1)
also given
a_{3}*a_{7} = 8
therefore (a + 2d)(a + 6d) = 8
rightarrow (3 - 4d + 2d)(3 - 4d + 6d) = 8
[using
(1)]
(3-2d)(3+2d)=8
rightarrow 9 - 4d ^ 2 = 8 rightarrow 4d ^ 2 = 1
rightarrow d ^ 2 = 1/4
rightarrow d = plus/minus 1/2
when d = 1 2
a = 3 - 4 * d = 3 - 4 * 1/2 = 3 - 2 = 1
when d
1
2
a = 3 - 4 * (d = 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.