Question

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

Answer 1

Answer:

$\text{Given that}\\\\\implies a_3 + a_7 = 6\\\\\implies a_3 \times a_7 = 8$

$\text{From this formula we get,} \\\\\implies a_3 = a + 2d \\\\\implies a_7 = a + 6d \\\\\implies a_3 + a_7 = 6 \\\\\implies a + 2d + a + 6d = 6\\\\\implies 2a + 8d = 6\\\\\text{Dividing by 2 throughout the equation we get,} \\\\\implies a + 4d = 3$

$\implies a = 3 - 4d \:\:\:\: \implies Equation \:\: 1$

$\implies( a + 2d ) ( a + 6d ) = 8\\\\\text{Substituting the value of 'a' we get,} \\\\\implies ( 3 - 4d + 2d ) ( 3 - 4d + 6d ) = 8\\\\\implies ( 3 - 2d )(3 + 2d ) = 8\\\\\implies 3^2 - (2d)^2= 8\\\\\implies 9 - 4d^2 = 8\\\\\implies 9 - 8 = 4d^2\\\\\implies 1 = 4d^2$

$\implies \dfrac{1}{4} = d^2 \\\\\text{Taking Square root on both sides }\\\\\implies \pm \dfrac{1}{2} = d$

$\text{Case 1: d =} + \dfrac{1}{2} \\\\\implies a = 3 - 4d \\\\\implies a = 3 - ( 4 \times \dfrac{1}{2} )\\\\\implies a = 3 - 2 \\\\\implies a = 1$

$\implies S_{20} = \dfrac{20}{2} \: [ 2 ( 1 ) + ( 20 - 1 ) \: \dfrac{1}{2} \: ]\\\\\implies S_{20} = 10\: [ 2 + \dfrac{19}{2} \: ]\\\\\implies S_{20} = 10 \: [ \: \dfrac{4+19}{2} \: ]\\\\\implies S_{20} = 10 \times \dfrac{23}{2} \implies 5 \times 23 \implies 115$

$\text{Case 2: d =} - \dfrac{1}{2} \\\\\implies a = 3 - 4d \\\\\implies a = 3 - ( 4 \times \dfrac{-1}{2} )\\\\\implies a = 3 - ( - 2 )\\\\\implies a = 3 + 2 \\\\\implies a = 5$

$\implies S_{20} = \dfrac{20}{2} \: [ \: 2 ( 5 ) + ( 20 - 1 ) \: \dfrac{-1}{2} \: ]\\\\\implies S_{20} = 10 \: [ \: 10 - \dfrac{19}{2} \: ]\\\\\implies S_{20} = 10 \: [\: \dfrac{20 - 19}{2} \: ]\\\\\implies S_{20} = 10 \times \dfrac{1}{2} \implies 5$

Hence when a is 1 and d is + 0.5, Sum is 115. When a is 5 and d is - 0.5, Sum is 5.

Answer 2

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