Answer
Verified
479.7k+ views
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.
Complete step-by-step answer:
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.
Hence, the answer is 11.
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.
Complete step-by-step answer:
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.
Hence, the answer is 11.
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.
Recently Updated Pages
Change the following sentences into negative and interrogative class 10 english CBSE
A Paragraph on Pollution in about 100-150 Words
One cusec is equal to how many liters class 8 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What were the social economic and political conditions class 10 social science CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
Trending doubts
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
How do you graph the function fx 4x class 9 maths CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
What is the past participle of wear Is it worn or class 10 english CBSE
Why did the British treat the Muslims harshly immediately class 10 social science CBSE
A Paragraph on Pollution in about 100-150 Words