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# Which term of the G.P. 5, 10, 20, 40, ..... is 5120 ?  Hint: Apply the basic formula for finding the nth terms of G.P. i.e. ${{T}_{n}}=a{{r}^{n-1}}$ where ‘a’ is the first term and ‘r’ is the common ratio of the G.P.

The given GP series is
5, 10, 20, 40, ............................(1)
We can define any GP series as
$A,AR,A{{R}^{2}},A{{R}^{3}}...........A{{R}^{n-1}}$
Where A = first term
R = common ratio
Here, next coming term is ‘R’ times of previous terms;
Hence, the common ratio is the ratio between any two continuous terms.
Now, series from equation (1) is;
5, 10, 20, 40..........
Hence, first term is 5 and common $Ratio=\dfrac{10}{5}=\dfrac{20}{10}=\dfrac{40}{20}=2$ from the basic definitions of geometric progression.
Therefore a = first term of given series = 5
And r = common ratio = 2.
Now, we have to determine which term will be 5120 in the given series.
As, we have general term or nth terms of G.P. is given in terms of first term and common ratio (r) as
${{a}_{n}}=a{{r}^{n-1}}...............\left( 2 \right)$
Now, we can observe from the question and equation (1) that term 5120 is a term of G.P. and we have to determine ‘n’ i.e. at what number it will come in series.
Hence, using equation (1), we have
\begin{align} & {{a}_{n}}=5120 \\ & r=2 \\ & a=5 \\ & n=? \\ \end{align}
Putting all the values to equation (1), we get
$5120=5{{\left( 2 \right)}^{n-1}}$
Transferring 5 to other side, we get
\begin{align} & {{\left( 2 \right)}^{n-1}}=\dfrac{5120}{5}=1024 \\ & {{\left( 2 \right)}^{n-1}}=1024 \\ \end{align}
Now, we can factorize 1024 and it can be written as ${{2}^{10}}$. Hence, above equation can be simplified as
${{\left( 2 \right)}^{n-1}}={{\left( 2 \right)}^{10}}$
As, bases of the surds are equal, hence power should also be equal to get the same result of the whole surd.
Hence,
\begin{align} & \left( n-1 \right)=10 \\ & n=11 \\ \end{align}
So, we get that 5120 be the 11th term of the given G.P.
Note: One can go wrong while applying ${{n}^{th}}$ term formula of G.P. One can put n = 5120 and try to calculate ${{a}_{n}}=a{{r}^{n-1}}$ which is wrong. So understanding the parameters used in ${{a}_{n}}=a{{r}^{n-1}}$is also an important side of the question.
One can go wrong by applying formula of ${{n}^{th}}$term of A.P i.e. ${{a}_{n}}=a+\left( n-1 \right)d$ which is wrong. As the given series is G.P. So, we cannot use any property of series A.P.
View Notes
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