Question

The sum of 7th and 3rd term of an AP is 6 and their product is . Find the sum of the first 20 terms of the ap

Answer 1

Here is your solution

Given:-

The sum of 3rd term and 7th term
=>a+2d+a+6d = 6
=>2a+8d = 6
After common taking
a + 4d = 3
a=3-4d ..................….......(1)

Their product is

(a+2d)×(a+6d) = 8 .........…(2)

Putting value of a = 3–4d in equation (2) to get

=>(a+2d)×(a+6d) = 8

=>(3–4d+2d)×(3–4d+6d) =8

=>(3–2d)×(3+2d) = 8

=>9–4d^2 = 8

=>4d^2 = 9-8

=>4d^2 = 1

=>d^2 = 1/4

=>d = 1/2.

From (1) a = 3–4d
a=> 3–4×1/2
a=>3-2
a=> 1

The sum of the first 20 terms

Sum of 16 terms = (n/2)[2a+(n-1)d]

= (20/2)[2×1 + (20–1)×1/2]

= 10[2+19/2]

= 10×23/2

= 230/2
=115

Hence,
The sum of first 20 term is 115

Hope it helps you

Answer 2

Answer:

Step-by-step explanation:

Here is your solution

Given:-

The sum of 3rd term and 7th term

=>a+2d+a+6d = 6

=>2a+8d = 6

After common taking

a + 4d = 3

a=3-4d ..................….......(1)

Their product is

(a+2d)×(a+6d) = 8 .........…(2)

Putting value of a = 3–4d in equation (2) to get

=>(a+2d)×(a+6d) = 8

=>(3–4d+2d)×(3–4d+6d) =8

=>(3–2d)×(3+2d) = 8

=>9–4d^2 = 8

=>4d^2 = 9-8

=>4d^2 = 1

=>d^2 = 1/4

=>d = 1/2.

From (1) a = 3–4d

a=> 3–4×1/2

a=>3-2

a=> 1

The sum of the first 16 terms

Sum of 16 terms = (20/2)[2a+(20-1)d]

= (20/2)[2×1 + (20–1)×1/2]

= 10[2+19/2]

= 10×23/2

= 230/2

=115

Hence,

The sum of first 20 term is 115

Hope it helps you