Question

in an A. P the sum of 3rd and 7th term is 6 and their product is 8. Find the first term and common difference of the A. P.

Answer 1

HELLO DEAR,

a​n​​=a+(n−1)d 

a = first term
n = no. of terms 
d = common difference

Given THAT:-

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   { putting a = 3 - 2d}

=>9 - 4d² = 8

=>d = +1/2 and - 1/2

when

d = +1/2

then

a = 3 - 2 = 1

and

when

d = -1/2  

a = 3 + 2 = 5

using formula for 

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

when d = +1/2

$S_{16}=\frac{16}{2}[2 \times 1+(16-1)d] \\ = > S_{16} = 8(2 + \frac{15}{2} ) \\ = > S_{n} = 4 \times 19 \\ = > S_{16} = 76$

WHEN, d=(-1/2)

$S_{16} = \frac{16}{2} (2 \times 5 + (16 - 1) \times ( - \frac{1}{2} )) \\ = > S_{16} = 8(10 - \frac{15}{2} ) \\ = > S_{16 }= 8 \times \frac{5}{2} \\ = > S_{16} = 4 \times 5 \\ = > S_{16} = 20$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

As we know that

an = a + ( n - 1 )d

Where

a = Fist term

n = Term

d = Common difference

an = AP on respective n Term

• Sum of third term and seventh term is 6

a3 + a7 = 6

a + 2d + a + 6d = 6

2a + 8d = 6

( dividing by 2 )

a + 4d = 3

a = 3 - 4d

• their product is 8

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

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

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

9 - 4d² = 8

- 4d² = 8 - 9

- 4d² = -1

d² = ± 1/2


When Common difference is 1/2

First term :-

a = 3 - 4d

a = 3 - 4 × 1/2

a = 3 - 2

a = 1

When common difference is - 1/2

a = 3 - 4d

a = 3 - 4 × - 1/2

a = 3 + 2

a = 5