Question

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.

Answer 1

We know that,

a_n = a + (n − 1) d

a₃ = a + (3 − 1) d

a₃ = a + 2d

Similarly, a₇ = a + 6d

Given that, a₃ + a₇ = 6

(a + 2d) + (a + 6d) = 6

2a + 8d = 6

a + 4d = 3

a = 3 − 4d                 ...(i)

Also, it is given that (a₃) × (a₇) = 8

⇒ (a + 2d) × (a + 6d) = 8

From equation (i),

(3 - 4d + 2d) × (3 - 4d + 6d) = 8

(3 - 2d) × (3 + 2d) = 8

9 - 4d² = 8

4d² = 9 - 8

4d² = 1 

d² =  `1/4`

d = `± 1/2`

d = `1/2 or 1/2`

From the equation (i)

`("When"  "d"  1/2)`

a = 3 - 4d

a = `3 -4(1/4)`

a = 3 - 2

a = 1

When  d  is − `1/2`

a = `3 - 4(-1/2)`

a = 3 + 2

a = 5

`S_n = n/2 [2a(n-1)s]`

(When a is 1 and d is `1/2`)

`S_16 = (16/2)[2"a" + (16 - 1)"d"]`

`S_16 = 8[2 xx 1 + 15(1/2)]`

`S_16 = 8(2 + 15/2)`

`S_16 = 8 xx 19/2`

S₁₆ = 76

(When a is 5 and d is `-1/2`)

a = `3 - 4(-1/2)`

a = `3 + 4/2`

a = 5

S₁₆ = `(16/2)[2a + (n - 1)d]`

= 8`[2(5) + 15(-1/2)]`

= `8[10 - 15/2]`

= `8 xx 5/2`

= 20