Question

The sum of the 3rd and the 7th terms of an AP is 6 and their product is 8. Find the sum of first 16th terms of the AP.

Answer 1

So we are given
[tex]t_{3} + t_{7} = 6 \\ t_{3} t_{7} = 8 \\ t_{n} = a+(n-1)d[/tex]
And we need to find
$S_{n} = \frac{n(2a+(n-1)d)}{2} \\ S_{16} = 8(2a+15d)$
Lets solve
Assume 
[tex] t_{5} = x[/tex]
then
[tex]t_{3} = x - 2d \\ t_{7} = x + 2d \\ t_{3} + t_{7} = 6 \\ 2x = 6 \\ x = 3[/tex]
Now
[tex]t_{3}t_{7} = 8 \\ (x-2d)(x+2d) = 8\\ 9 - 4d^2 = 8 \\ d^2 = \frac{1}{4} \\ d = + \frac{1}{2},- \frac{1}{2} [/tex]
now
$t_{5} = x = a +4d = 3$
So a can be
$a = 3+2,3-2$
$a = 5, 1$
2 possible A.P  are 
Now 
$S_{16} = 8(2a+15d) \\ a = 5, d = \frac{-1}{2}, S_{16} = 20 \\ a = 1, d = \frac{1}{2}, S_{16} = 76$
Proof:
for first A P
$t_{3} = 5 - \frac{2}{2} = 4 \\ t_{7} = 5 - \frac{6}{2} =2 \\ t_3 + t_7 = 6, \\ t_3t_9 = 8$
For second AP
$t_{3} = 1 + \frac{2}{2} = 2 \\ t_{7} = 1 + \frac{6}{2} =4 \\ t_3 + t_7 = 6, \\ t_3t_9 = 8$

Answer 2

3rd term = a + 2d
7th term = a + 6d

ATQ,
      
        a + 2d + a + 6d = 6
        2a + 8d             = 6
        2 (a + 4d)          = 6
        a + 4d               = 6/2
        a + 4d               = 3
        a                       = 3 - 4d -------------1

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

1 in 2 ⇒

(3 - 4d + 2d)(3 -4d +6d) = 8
(3 - 2d)(3 + 2d)              = 8
3² - (2d)²                        = 8
9 - 4d²                            = 8
4d²                                 = 1
d²                                   = 1/4
d                                    = √1/4
                                      = +1/2  or  -1/2

taking d = +1/2
put in equation 1 ⇒
a   = 3 - 4 × 1/2
a   = 1

taking d = -1/2
put in equation 1 ⇒
a   = 3 - 4 × -1/2
a   = 5

when a = 1 and d = +1/2
Sum of 16 terms  =  16/2(2×1 + 15×1/2) = 76

when a = 5 and d = -1/2
sum of 16 terms   =  16/2(2×5 + 15×-1/2)= 20