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

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

So by first condition we get t3+t7=(a+2d)+(a+6d)= 6

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

By 2nd condition we get t3xt7=(a+2d)x(a+6d)= 8

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

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



=> 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

S16= 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

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