The sum of third and seventeenth terms of an
a.P in 6 and their product is 8. Find the sum of first sixteenth terms of an
a.P
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a3+a17=6
a+2d+a+16d=6
2a+18d=6
a+9d=3
a=3-9d........1
a3×a17=8
(a+2d)(a+16d)=8
put in a value.........1 for
(3-9d+2d)(3-9d+16d)=8
(3-7d)(3+7d)=8 [(a+b)(a-b)=a^2-b^2]
(3)^2-(7d)^2=8
9-49d^2=8
-49d^2=8-9
d^2=-1/-49
d^2=(1/7)^2
d=1/7
a=3-9d=3-9×1/7=(21-9)/7=12/7
sn=n/2[2a+(n-1)d]
s16=16/2[2×12/7+15×1/7]
=8(24/7+15/7)
=8(24+15)/7
=8×39/7
=312/7
a+2d+a+16d=6
2a+18d=6
a+9d=3
a=3-9d........1
a3×a17=8
(a+2d)(a+16d)=8
put in a value.........1 for
(3-9d+2d)(3-9d+16d)=8
(3-7d)(3+7d)=8 [(a+b)(a-b)=a^2-b^2]
(3)^2-(7d)^2=8
9-49d^2=8
-49d^2=8-9
d^2=-1/-49
d^2=(1/7)^2
d=1/7
a=3-9d=3-9×1/7=(21-9)/7=12/7
sn=n/2[2a+(n-1)d]
s16=16/2[2×12/7+15×1/7]
=8(24/7+15/7)
=8(24+15)/7
=8×39/7
=312/7
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