Math, asked by Sathvikng, 1 year ago

The sum of third and seventh terms of an AP is 6 and thier product is 8. Find the sum of first 16 terms of the AP

Answers

Answered by ssourabh
1
actually it has two results 76 & -20
Attachments:
Answered by Anonymous
0

\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}}}



Similar questions