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Sum of an infinite G.P. is 2 and sum of their cubes is 24, then 5th term of the G.P. is
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a
3/16
b
3/8
c
– 3/8
d
– 3/16
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Correct option is A
a1−r=2,a31−r3=24∴8(1−r)31−r3=24⇒1−2r+r21+r+r2=3⇒2r2+5r+2=0⇒r=−2,−1/2As |r|<1,r=−1/2,a=3Now, t5=ar4=3/16Similar Questions
If and and , then
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