Question

The sum of third and seventh terms of an A.P. is 6 and their product is 8. Find the sum of first 16th terms of the A.P.

Answer 1

Hi

Here is your answer

Given

Third term be a₃

Seventh term be a₇

Sum of third and seventh term = 6

a₃ + a₇ = 6

Here n = 3 and n = 7

a + ( 3 - 1) d + a + ( 7 - 1 ) d = 6

a + 2d + a + 6d = 6

2a + 8d = 6

Divide the above equation with 2,

a + 4d = 3  

a = 3 - 4d ---- (1)

Product of third and seventh term = 8

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

Substitute a value in the equation,

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

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

9 - 4d² = 8

-4d² = 8 - 9

4d² = 1

d² = 1/4

d = + 1/2 or -1/2

By substituting d = +1/2 in a = 3 - 4d

a = 3 - 4/2

a = 3 - 2

a = 1

===

d = -1/2 in a = 3-4d

a = 3 - (-4/2)

a = 3+2

a = 5

===

Now we have to find Sum of 16 terms,

Sn = n/2[2a+(n-1)d]

n = 16 , a = 1, d = 1/2

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

Sn = 8 [ 2 + 15/2 ]

Sn = 8 [ 4 + 15]/2

Sn = 4(19)

Sn = 76..

Then, n = 16, a = 5, d = -1/2

Sn = n/2[2a+(n-1)d]

Sn = 16/2 [ 2*5 + (16-1) ( - 1/2) ]

Sn = 8 [ 10 - 15/2 ]

Sn = 8 [20 - 15]/2

Sn = 4 [ 5]

Sn = 20

=========================

Hope it helps you....

Answer 2

$\bf\huge\boxed{\boxed{\bf\huge\:Hello\:Mate}}}$

$\bf\huge Let\: the\: AP\: be\: a - 4d , a - 3d , a - 2d , a - d , a , a + d , a + 2d , a + 3 d$

$\bf\huge => a_{3} = a - 2d$

$\bf\huge => a_{7} = a - 2d$

$\bf\huge => a_{3} + a_{7} = a - 2d + a - 2d = 6$

$\bf\huge => 2a = 6$

$\bf\huge => a = 3 (Eqn 1)$

$\bf\huge Hence\: (a - 2d) (a + 2d) = 8$

$\bf\huge => a^2 - 4d^2 = 8$

$\bf\huge => 4d^2 = a^2 - 8$

$\bf\huge => 4d^2 = (3)^2 - 8 = 9 - 8 = 1$

$\bf\huge => d^2 = \frac{1}{4}$

$\bf\huge => d = \frac{1}{2}$

$\bf\huge\texttt Hence$

$\bf\huge S_{16} = \frac{16}{2} [2\times (a - 4d)+ (16 - 1)\times d]$

$\bf\huge => 8[2\times (3 - 4\times \frac{1}{2})+ 15\times \frac{1}{2}]$

$\bf\huge => 8[2 + \frac{15}{2}]= 8\times \frac{19}{2} = 76$

$\bf\huge => d = - \frac{1}{2}$

$\bf\huge Putting\:the\: Value\: of\: D$

$\bf\huge S_{16} = \frac{16}{2} [2\times (a - 4d)+ (16 - 1)\times d]$

$\bf\huge => 8[2\times (3 - 4\times - \frac{1}{2})+ 15\times - \frac{1}{2}]$

$\bf\huge => 8[2\times 5 - \frac{15}{2}]$

$\bf\huge => 8 [ \frac{20 - 15}{2}]$

$\bf\huge => 8\times \frac{5}{2} = 20$

$\bf\huge Hence$

$\bf\huge S_{16} = 20 , 76$

$\bf\huge\boxed{\boxed{\:Regards=\:Yash\:Raj}}}$