Question

the sum third and seventh term if an appointment is 6 and their product is 8 find the sum of first sixteen terms of ap

Answer 1

3rd term=a+2d
7th term=a+6d
given
sum of 3rd and 7th term =a+2d+a+6d=2a+8d
=2a= -8d
=a= -8d/2= -4d ..........1
And product of 3rd and 7th term=(a+2d)(a+6d)
=a²+6ad+2ad+12d²=a²+8ad+12d²
Putting the value of a
=(-4)²+8×-4×d+12d²
=16-32d+12d²
=12d²-32d+16
dividing by 4
=3d²-8d+4=0
Using spitting middle term
=3d²-6d-2d+4=0
=3d(d-2)-2(d-2)=0
=(3d-2)(d-2)=0
3d-2=0 d-2=0
d=2/3 or 2
taking d=2
putting in equation 1
a= -4×2= -8
so, First sixteen terms are
-8,-6,-4,-2,0,2,4,6,8,10,12,14,16,18,20,22

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf\huge\texttt Hence$

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

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

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

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

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

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

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

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

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

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

$\bf\huge Hence$

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

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