Let the first term of AP be **a **and its common difference be **d.**

So by first condition we get t_{3}+t_{7}=(a+2d)+(a+6d)= 6

=> a= 3-4d............[1]

By 2nd condition we get t_{3}xt_{7}=(a+2d)x(a+6d)= 8

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

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

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

=>d^{2}=1/4

**=> d=+0.5 and -0.5**

When d = 0.5 then a= 1

and when d= -0.5 then a= 5

**when a=1 and d =0.5 the sum of first 16 terms will be **

**S**_{16}= 16/2[2x1+(16-1)x0.5]=76

**when a=5 and d =-0.5 the sum of first 16 terms will be**

**S'**_{16}= 16/2[2x5-(16-1)x0.5]=20