Question

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

Answer 1

Hint: It is given that the sequence is in geometric progression (G.P.), The sum of $n$ terms of a G.P is given by the formula $S_n={\displaystyle \frac{a(1-r^n)}{(1-r)}}$, where $a$is the first term of the sequence and $r$ is the common ratio. Using the first three terms of the given G.P we will find the value of $a$ and $r$, substituting these values in the sum formula we can find the value of ‘n’.

Complete step-by-step solution:
Geometric progression is a sequence of non-zero numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. The general form of a geometric progression is given by $a,ar,a{r}^2,a{r}^3,...$ where $a$ the first term of the sequence and $r$ is the common ratio. ‘r’ can be calculated by dividing the succeeding term by the preceding term i.e., $r={\displaystyle \frac{ar}{a}}={\displaystyle \frac{a{r}^2}{ar}}$.
Comparing the given sequence $3,{\displaystyle \frac{3}{2}},{\displaystyle \frac{3}{4}},...$ with standard form the values of ‘a and ‘r’ are,
The first term $a=3$ and the common ratio $r={\displaystyle \frac{\left[{\displaystyle \frac{3}{2}}\right]}{3}}={\displaystyle \frac{3}{2}}\times {\displaystyle \frac{1}{3}}={\displaystyle \frac{1}{2}}$.
Also they have given the sum of ‘n’ terms , $S_n={\displaystyle \frac{3069}{512}}$, now we will substitute all these obtained values in the formula for sum of ‘n’ terms of a G.P , $S_n={\displaystyle \frac{a(1-r^n)}{(1-r)}}\Rightarrow {\displaystyle \frac{3069}{512}}={\displaystyle \frac{3\left(1-{\left({\displaystyle \frac{1}{2}}\right)}^n\right)}{\left(1-{\displaystyle \frac{1}{2}}\right)}}={\displaystyle \frac{3\left(1-{\left({\displaystyle \frac{1}{2}}\right)}^n\right)}{\left({\displaystyle \frac{1}{2}}\right)}}=3\times 2\left(1-{\left({\displaystyle \frac{1}{2}}\right)}^n\right)$ ,on cross multiplication we get, $\begin{array}{rl}& \Rightarrow \left(1-{\left({\displaystyle \frac{1}{2}}\right)}^n\right)={\displaystyle \frac{3069}{512\times 6}}={\displaystyle \frac{1023}{1024}}\\ & \Rightarrow {\left({\displaystyle \frac{1}{2}}\right)}^n=1-{\displaystyle \frac{1023}{1024}}={\displaystyle \frac{1024-1023}{1024}}={\displaystyle \frac{1}{1024}}={\displaystyle \frac{1}{2^{10}}}\end{array}$
$\Rightarrow {\left({\displaystyle \frac{1}{2}}\right)}^n={\left({\displaystyle \frac{1}{2}}\right)}^{10}$, since the base value is same equating the power values we get the value of $n=10.$
Therefore $10$ terms of the G.P. $3,{\displaystyle \frac{3}{2}},{\displaystyle \frac{3}{4}},...$ are needed to give the sum ${\displaystyle \frac{3069}{512}}$.

Note: Any progression is a special type of sequence in which it is possible to obtain a formula for the general term of the given sequence. Based on the general term there are different types of progressions out of which Arithmetic progression (A.P) and geometric progression are commonly used. In A.P. the terms can be obtained by adding or subtracting a constant to the preceding term, wherein in case of geometric progression each term is obtained by multiplying or dividing a constant to the preceding term.