Please help me solving the sum!!

How many terms of G.P 3 , 3/2 , 3/4... must be taken to give the sum 3069/5122 ???

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Solution

This question is not correctly given because of sum of first three terms of this GP 3 , 3/2, 3/4 = 5.25 but in this question sum of this sequence is given as 3069/5122= 0.599 which is even less than 1. the sum of this Geometric sequence 3 , 3/2 , 3/4... must be greater than 5.25
so how can this possible ?

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Similar questions

G . P .3, \frac{3}{2}, \frac{3}{4} . . . . . . .

\frac{3069}{512}

How many terms of the G.P. $3, \frac{3}{2}, \frac{3}{4} . . . . .$ are needed to give the sum $\frac{3069}{512}$ ?

3, \frac{3}{2}, \frac{3}{4} . . .,

\frac{3069}{512} ?

How many terms of the G.P. 3, $\frac{3}{2}, \frac{3}{4}, . . . .$ be taken together to make $\frac{3069}{512}$ ?

3, \frac{3}{2}, \frac{3}{4},, .

\frac{3069}{512}

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