Question

in an ap if the sum of the third term and seventh term is 6and their product is 8 find the sum of first 16terms
​

Answer 1

Answer:-

Given:

Sum of 3rd & 7th term of an AP = 6.

We know that,

nth term of an AP = a + (n - 1)d

Hence,

→ 3rd term + 7th term = 6

→ a + (3 - 1)d + a + (7 - 1)d = 6

→ 2a + 2d + 6d = 6

→ 2(a + d + 3d) = 6

→ a + 4d = 6/2

→ a + 4d = 3

→ a = 3 - 4d -- equation (1)

And,

Also given that,

Their product = 8

→ (a + 2d) * (a + 6d) = 8

Substitute the value of a from equation (1).

→ (3 - 4d + 2d) * (3 - 4d + 6d) = 8

→ (3 - 2d) * (3 + 2d) = 8

Using (a + b) * (a - b) = a² - b² we get,

→ 3² - (2d)² = 8

→ 9 - 8 = 4d²

→ 1/4 = d²

→ 1/2 = d

Putting the value of d in equation (1) we get,

→ a = 3 - 4 * 1/2

→ a = 3 - 2

→ a = 1

We know,

Sum of first n terms of an AP – S(n) = n/2 * [ 2a + (n - 1)d ]

→ S(16) = 16/2 * [ 2(1) + (16 - 1) * (1/2) ]

→ S(16) = 8 ( 2 + 15/2 )

→ S(16) = 8 [ (4 + 15) / 2 ]

→ S(16) = 4 * 19

→ S(16) = 76

Therefore, the sum of first 16 terms of the given AP is 76.

Answer 2

Answer:

hey mate mark me brainliest and give a thanks

Step-by-step explanation:

so using formula for n th term  

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  

when d = +1/2

Similarly when d = -1/2