
Find the $10^{th}$ term & $n^{th}$ term of the G.P. 5, 25, 125 and so on.
Answer
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Hint: For the given Geometric Progression (GP) , we have the first term (a=5) with the common ratio (r=5) and the $n^{th}$ term of the GP being given by the formula: \[{{T}_{n}}=a{{r}^{n-1}}\]. So we just have to replace the value to get the \[{{n}^{th}}\] term and the \[{{10}^{th}}\]term of the GP.
Complete step by step solution:
In the question, we have to find the\[{{10}^{th}}\]term & \[{{n}^{th}}\] term of the G.P. 5, 25, 125. So, to find any term of the progression, we will apply the formula for the $n^{th}$ term of the given GP. Now GP is a geometric progression that has the common ratio between the two consecutive terms. So here the common ratio will be found by finding the ratio of second term to the first term. The common ratio (r) is given as:
\[\begin{align}
& \Rightarrow r=\dfrac{25}{5} \\
& \Rightarrow r=5 \\
\end{align}\]
So here we got that the common ratio is 5. Now, we need to identify the first term of this G.P. 5, 25, 125… which is written as a and it is a=5. So the \[{{n}^{th}}\] term of this GP will be given by the formula:
\[{{T}_{n}}=a{{r}^{n-1}}\]. Here putting the values of a=5, r=5, we will get the $n^{th}$ term, as shown below:
\[\begin{align}
& \Rightarrow {{T}_{n}}=a{{r}^{n-1}} \\
& \Rightarrow {{T}_{n}}=5\times {{5}^{n-1}} \\
& \Rightarrow {{T}_{n}}={{5}^{n}} \\
\end{align}\]
So the expression of the \[{{n}^{th}}\] term is \[{{T}_{n}}={{5}^{n}}\]. Next, using this we can find the \[{{10}^{th}}\] term.
So, we just need to put n=10 in the other expression to get the $10^{th}$ term. That will be as follows:
\[\begin{align}
& \Rightarrow {{T}_{n}}={{5}^{n}} \\
& \Rightarrow {{T}_{10}}={{5}^{10}} \\
\end{align}\]
Thus the \[{{10}^{th}}\] of the G.P. 5, 25, 125 and so on will be \[{{5}^{10}}\]. This is the required answer.
Note: Whenever we are given a sequence, we always have to first check if the sequence is in Arithmetic progression (AP) or in Geometric progression (GP). The GP will have the common ratio and the AP will have the common difference between the two consecutive terms.
Complete step by step solution:
In the question, we have to find the\[{{10}^{th}}\]term & \[{{n}^{th}}\] term of the G.P. 5, 25, 125. So, to find any term of the progression, we will apply the formula for the $n^{th}$ term of the given GP. Now GP is a geometric progression that has the common ratio between the two consecutive terms. So here the common ratio will be found by finding the ratio of second term to the first term. The common ratio (r) is given as:
\[\begin{align}
& \Rightarrow r=\dfrac{25}{5} \\
& \Rightarrow r=5 \\
\end{align}\]
So here we got that the common ratio is 5. Now, we need to identify the first term of this G.P. 5, 25, 125… which is written as a and it is a=5. So the \[{{n}^{th}}\] term of this GP will be given by the formula:
\[{{T}_{n}}=a{{r}^{n-1}}\]. Here putting the values of a=5, r=5, we will get the $n^{th}$ term, as shown below:
\[\begin{align}
& \Rightarrow {{T}_{n}}=a{{r}^{n-1}} \\
& \Rightarrow {{T}_{n}}=5\times {{5}^{n-1}} \\
& \Rightarrow {{T}_{n}}={{5}^{n}} \\
\end{align}\]
So the expression of the \[{{n}^{th}}\] term is \[{{T}_{n}}={{5}^{n}}\]. Next, using this we can find the \[{{10}^{th}}\] term.
So, we just need to put n=10 in the other expression to get the $10^{th}$ term. That will be as follows:
\[\begin{align}
& \Rightarrow {{T}_{n}}={{5}^{n}} \\
& \Rightarrow {{T}_{10}}={{5}^{10}} \\
\end{align}\]
Thus the \[{{10}^{th}}\] of the G.P. 5, 25, 125 and so on will be \[{{5}^{10}}\]. This is the required answer.
Note: Whenever we are given a sequence, we always have to first check if the sequence is in Arithmetic progression (AP) or in Geometric progression (GP). The GP will have the common ratio and the AP will have the common difference between the two consecutive terms.
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