The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.
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Answer:
a+2d+a+6d=6—1
(a+2d)(a+6d)=8—2
2a+8d=6
a^2+12d^2+8ad=8
2a=6-8d
a=6-8d/2
(6-8d/2)^2+12d^2+8(6-8d/2)d=8
36-64d/4+12d^2+(48-64d/16)d=8
144-256d+72d^2+48d-64d^2=128
144-208d+8d^2=128
8d^2-208d+144-128=0
d^2-26d+2=0
d=26
2a+8(26)=6
2a+208=6
2a=6-208
2a=-202
a=-101
For the 16th term
a+15d
-101+15(26)
=289
Step-by-step explanation:
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