Question

The third term of a G.P is 27 and sixth term is 8 ; find the sum to infinite terms.
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ANSWER = a=243/4 , r=2/3 , S=729/4

Answer 1

any solution for this one?

Answer 2

let the terms are t1,t2,t3,t4,t5...........................
for G.P series, a is the first term and r is the common ratio=t2/t1.
here,
n th term tn=ar^(n-1).
so from the given data,
t3=ar^(3-1)=27 and t6=ar^(6-1)=8
so t6/t3=ar^5/ar^2
=>8/27=r^(5-2)
=>(2/3)^3=r^3
=> r=2/3

Answer 3

thanks for the r

Answer 4

again t3=ar^2
=>a=t3/r^2=27/(2/3)^2=27*9/4=243/4

Answer 5

and how do i get a and S

Answer 6

now the sum for infinite series with r<1 is,

s=a/(1-r)=(243/4)/(1-2/3)=(243/4)/(1/3)=243*3/4=729/4