Question

The sum of third and seventh term of an A.P is 6 and their product is 8. find the sum of 16th term of the A.P​

Answer 1

Answer:

4,2

Step-by-step explanation:

a3 +a7=6

a3×a7=8

this may be helpful for you

Answer 2

Answer: 20

Step-by-step explanation: Using formula for nth term, $a_{n} = 1 + (n - 1)d$

a = first term

n = no. of terms 

d = common difference

So given,

a + (3-1)d + a (7-1)d = 6

2a + 8d = 6

a + 4d = 3

a = 3 - 4d

(a + 2d)(a + 6d) = 8 

(3 - 2d)( 3 + 2d) = 8   {substituting a = 3 - 2d}

9 - 4d² = 8

d = +1/2 and - 1/2

So when d = +1/2 then a = 3 - 2 = 1 and when d = -1/2  a = 3 + 2 = 5

So using formula for $S_{n} = \frac{n}{2} [2a + (n - 1)d]$

when d = 1/2

$S_{n} = \frac{16}{2} [2 * 1 + (16 - 1)d]$

$S_{n} = 76$

When d = +1/2,

$S_{n} = 20$