Question

merry christmas guysss my question is on maths......The sum of the thrid and the seventh terms of an AP is 6 and their product is 8. find the sum of first sixteen terms of the AP

Answer 1

Answer:

104

Step-by-step explanation:

merry christmas

third term = a3 = a+2d

seventh term = a7 = a+6d

the sum is 6 and product is 8

first lets find the sum

a+2d+a+6d=6

2a+8d=6

a+4d=3 ( dividing the whole equation by 2 to simplify)

a=3-4d

now lets find the product

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

putting a=3-4d in this equation

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

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

$3^{2} - (2d)^{2}$=8

9-4$d^{2}$=8

-$d^{2}$= 8-9 = -1

d = 1

a= 3-4d=3-4= -1

now lets find the 16th term

a16=a+15d=-1+15=14

Sn=n/2(a+an)

=16/2(-1+14)

=8(13)=104

hope this ans was helpful

feel free to clear your doubts

pls mark as brainliest

thank you

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf\huge\texttt Hence$

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

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

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

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

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

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

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

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

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

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

$\bf\huge Hence$

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

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