Question

If the second and fifth terms of a GP are \[24\]and \[3\] respectively, then the sum of the first sixth terms is
(1) \[181\]
 (2) \[\dfrac{181}{2}\]
(3) \[189\]
(4) \[\dfrac{189}{2}\]

Answer 1

Hint: To find the sum of any terms of the GP, first we should know about the first term and the common ratio of the GP. If the first term and the common ratio are not given then we will find both of them with the help of the given equations. After knowing the first term and common ratio we will put them in the respective formula and our answer will be obtained.

Complete step by step answer:
Since we know that, \[{{n}^{th}}\] term of the G.P. is represented as \[a{{r}^{n-1}}\].
Where \[a\] is the first term and \[r\] is the common ratio. Common ratio is calculated in G.P. series by dividing the second term with the first term.
Suppose,
First term =\[a\]
Second term =\[ar\]
Now after dividing the second term with the first term, a common ratio will be obtained.
Now according to our question it is given that,
Second term of the GP \[=24\]
Fifth term of the GP \[=3\]
So,
Second term can be written as
\[ar=24\]…………. (1)
Fifth term can be written as
\[a{{r}^{4}}=3\]………….. (2)
We have to find the sum of the first six terms of the GP, so for that we have to find the first term and common ratio of the GP .So common ratio is calculated by dividing equation (2) with equation (1) which is as follows,
\[\dfrac{a{{r}^{4}}}{ar}=\dfrac{3}{24}\]
\[\Rightarrow {{r}^{3}}=\dfrac{1}{8}\]
\[\Rightarrow {{r}^{3}}=\dfrac{1}{{{(2)}^{3}}}\]
\[\Rightarrow r=\dfrac{1}{2}\]
So the common ratio is \[\dfrac{1}{2}\].
Now with the help of this we will find the first term by putting this value in equation (1).We get,
\[\begin{align}
  & ar=24 \\
 & \Rightarrow a\times \dfrac{1}{2}=24 \\
\end{align}\]
\[a=48\]
So the first term of the G.P. is \[48\]. Now we will find the sum of the six terms of the GP.
We know that,
Sum of \[n\] terms of G.P. can be calculated by the formula,
\[{{S}_{n}}=\left\{ \begin{align}
  & \dfrac{a(1-{{r}^{n}})}{1-r},r<1 \\
 & \dfrac{a({{r}^{n}}-1)}{r-1},r>1 \\
\end{align} \right\}\]
Where,
\[r=\] Common Ratio
\[a=\] First term
\[n=\] Number of terms
As we know that,
\[r=\dfrac{1}{2}\] , Which is less than \[1\]
\[a=48\]
\[n=6\]

So we apply the formula for sum, we get
 \[{{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r}\]
Put the values of all variables in above equation we get,
\[\Rightarrow {{S}_{6}}=\dfrac{48\left( 1-{{\left( \dfrac{1}{2} \right)}^{6}} \right)}{1-\dfrac{1}{2}}\]
\[\Rightarrow {{S}_{6}}=\dfrac{48\left( 1-\dfrac{1}{64} \right)}{\dfrac{2-1}{2}}\]
\[\Rightarrow {{S}_{6}}=\dfrac{48\left( \dfrac{63}{64} \right)}{\dfrac{1}{2}}\]
\[\Rightarrow {{S}_{6}}=\dfrac{48\times 2\times 63}{64}\]
\[\Rightarrow {{S}_{6}}=\dfrac{189}{2}\]
 So the sum of the first six terms of G.P. will be , \[{{S}_{6}}=\dfrac{189}{2}\]

So, the correct answer is “Option 4”.

Note: Other than G.P. series we have two other series Arithmetic progression and Harmonic progression. In Arithmetic progression we have a common difference in which every two neighboring numbers have the same difference and harmonic progression is reciprocal to arithmetic progression .