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# How many terms of the G.P. $3,\dfrac{3}{2},\dfrac{3}{4},...$ are needed to give the sum $\dfrac{{3069}}{{512}}$? Verified
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 ${S_n} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}}$, 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’.
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., $r = \dfrac{{ar}}{a} = \dfrac{{a{r^2}}}{{ar}}$.
Comparing the given sequence $3,\dfrac{3}{2},\dfrac{3}{4},...$ with standard form the values of ‘a and ‘r’ are,
The first term $a = 3$ and the common ratio $r = \dfrac{{\left[ {\dfrac{3}{2}} \right]}}{3} = \dfrac{3}{2} \times \dfrac{1}{3} = \dfrac{1}{2}$.
Also they have given the sum of ‘n’ terms , ${S_n} = \dfrac{{3069}}{{512}}$, now we will substitute all these obtained values in the formula for sum of ‘n’ terms of a G.P , ${S_n} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}} \Rightarrow \dfrac{{3069}}{{512}} = \dfrac{{3\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^n}} \right)}}{{\left( {1 - \dfrac{1}{2}} \right)}} = \dfrac{{3\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^n}} \right)}}{{\left( {\dfrac{1}{2}} \right)}} = 3 \times 2\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^n}} \right)$ ,on cross multiplication we get, \eqalign{ & \Rightarrow \left( {1 - {{\left( {\dfrac{1}{2}} \right)}^n}} \right) = \dfrac{{3069}}{{512 \times 6}} = \dfrac{{1023}}{{1024}} \cr & \Rightarrow {\left( {\dfrac{1}{2}} \right)^n} = 1 - \dfrac{{1023}}{{1024}} = \dfrac{{1024 - 1023}}{{1024}} = \dfrac{1}{{1024}} = \dfrac{1}{{{2^{10}}}} \cr}
$\Rightarrow {\left( {\dfrac{1}{2}} \right)^n} = {\left( {\dfrac{1}{2}} \right)^{10}}$, since the base value is same equating the power values we get the value of $n = 10.$
Therefore $10$ terms of the G.P. $3,\dfrac{3}{2},\dfrac{3}{4},...$ are needed to give the sum $\dfrac{{3069}}{{512}}$.