From the given statements, we can write,

a_{3 +} a₇ = 6 ……………….(i)

a₃ ×a₇ = 8  ………………..(ii)

By the nth term formula,

a_n = a+(n−1)d

Third term, a₃ = a+(3 -1)d

a₃ = a + 2d  ………………(iii)

And Seventh term, a₇ = a+(7-1)d

a₇ = a + 6d ………………..(iv)

From equation (iii) and (iv), putting in equation (i), we get,

a+2d +a+6d = 6

2a+8d = 6

a+4d=3

or

a = 3–4d  ……………(v)

Again putting the eq.(iii) and (iv), in eq. (ii), we get,

(a+2d)×(a+6d) = 8

Putting the value of a from equation (v), we get,

(3–4d +2d)×(3–4d+6d) = 8

(3 –2d)×(3+2d) = 8

3² – 2d² = 8

9 – 4d² = 8

4d² = 1

d = 1/2 or -1/2

Now, by putting both the values of d, we get,

a = 3 – 4d = 3 – 4(1/2)

= 3 – 2 = 1, when d = 1/2

a = 3 – 4d = 3 – 4(-1/2)

= 3+2 = 5, when d = -1/2

S_n = n/2 [2a +(n – 1)d]

So, when a = 1 and d=1/2

Then, the sum of first 16 terms are;

S₁₆ = 16/2 [2 +(16-1)1/2]

= 8(2+15/2) = 76

And when a = 5 and d= -1/2

Then, the sum of first 16 terms are;

S₁₆ = 16/2 [2.5+(16-1)(-1/2)]

= 8(5/2)=20