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rafimallickgals
@rafimallickgals
June 2021
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Find the third term of a GP whose common ratio is 3 and the sum whose first seven term is 2186.
Answers & Comments
niraj449170
Given;
r=3;and S7=2186
S7=a(1 - 3(to the power 7)) / 1-3
2186=a(1 - 2187) / -2
2186 * -2=a(-2186)
-4372=a(-2186)
-4372/-2186=a
a=2.
T3=ar^2
=2*(3)^2
=18
therefore the third term is : 18
7 votes
Thanks 6
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Answers & Comments
r=3;and S7=2186
S7=a(1 - 3(to the power 7)) / 1-3
2186=a(1 - 2187) / -2
2186 * -2=a(-2186)
-4372=a(-2186)
-4372/-2186=a
a=2.
T3=ar^2
=2*(3)^2
=18
therefore the third term is : 18