How many terms of the G.P. $$3,3 / 2,3 / 4$$ are needed to give the sum $$3069 / 512 ?$$
From the question, it is given that,
Sum of the terms $$S_{n}=3069 / 512$$
First term $$a=3$$
Common ratio $$r=(3 / 2) / 3$$
$$=(3 / 2) \times(1 / 3)\\$$
$$= 1 / 2$$
We know that,
$$S_{n}=a\left(1-r^{n}\right) /(1-r)$$
$$(3069 / 512)=3\left[1-(1 / 2)^{n}\right] /(1-1 / 2)$$
$$(3069 / 512)=(2 \times 3)\left[1-(1 / 2)^{n}\right]$$
$$1-(1 / 2)^{n}=3069 /(512 \times 6)$$
$$1-(1 / 2)^{n}=1023 / 1024$$
$$(1 / 2)^{n}=1-(1023 / 1024)$$
$$(1 / 2)^{n}=(1024-1023) / 1024$$
$$(1 / 2)^{n}=1 / 1024$$
$$(1 / 2)^{n}=(1 / 2)^{10}$$
By comparing both LHS and RHS, $$n=10$$
Therefore, there are 10 terms are in the G.P.