Math, asked by ravindranathanmp, 1 year ago

The sum of 7terms and 3rd term of an AP is 6 and their product is 8.find the sum of first 20th terms

Answers

Answered by Anonymous
2
The sum of 7th term and 3rd term of an AP

we know that nth Term in AP = a + (n-1)d

7th Term = a +(7-1)d = a + 6d...(1)

3rd Term= a + (3 - 1)d = a + 2d...(2)

Given ''sum of (1) & (2) is 6

(a +6d) + (a + 2d) = 6

(2a + 8d) = 6

2(a + 4d) = 2(3)

a + 4d = 3....(3)

a = 3 - 4d....(4)


Given"product of (1) & 2 is 8"

(a + 6d) (a +2d) = 8

substitute (4)

(3 -4d + 6d) ( 3 -4d + 2d) = 8

(3 +2d) ( 3 -2d) = 8

its in the form of (a+b)(a-b) = (a^2 - b^2)

(3)^2 - (2d)^2 = 8

9 - 4d^2 = 8

9-8 = 4d^2

1/4 = d^2

√(1/4) = d

1/2 = d... (5) substitute it in 4

a = 3 -4d

a = 3 - 4(1/2)

a = 3 - 2

a = 1... (6)


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

sum of 20 terms

= 20/2 [2(1) + (20 - 1)1/2]

= 10[ 2 + 19/2]

= 10[2 + 9.5]

= 10[11.5]

= 115

Answered by SmãrtyMohït
4
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
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