Question

The product of three consecutive terms of a G.P is 512. If 4 is added to first and second of these terms, then the three now form an A.P. Then the sum of original three terms of the given G.P is
$\begin{aligned} & \left( A \right)36 \\\ & \left( B \right)24 \\\ & \left( C \right)32 \\\ & \left( D \right)28 \\\ \end{aligned}$

Answer 1

We start solving this problem by going through the definition of G.P and A.P and then we assume the three terms of G.P as $\dfrac{a}{r}$, $a$, and $ar$. Then we find the product of the terms and equate the result to 512 to find the value of $a$. Then we add 4 to the first and second terms and we find the common difference of first and second term, second and third term, and equate them. By solving them we can find the value of $a$. Then we substitute the value in the original G.P and find the sum of those terms.

Complete step-by-step solution:
A sequence of numbers is said to be in Geometric Progression if each term in the sequence differs by its next term by a common ratio. We denote a geometric progression as $a,ar,a{{r}^{2}},a{{r}^{3}},........$ here a is the first term and r is the common ratio.
A sequence of numbers is said to be in Arithmetic Progression if each term in the sequence differs by its next term by a common difference. We denote an arithmetic progression as $a,a+d,a+2d,a+3d,........$ here a is the first term and d is a common difference.
We were given that product of three terms in G.P is 512.
Let us consider the three terms as $\dfrac{a}{r}$, $a$ and $ar$. Here r is common ratio.
So now let us multiply them and equate it to 512.
$\begin{aligned} & \Rightarrow \dfrac{a}{r}\times a\times ar=512 \\\ & \Rightarrow {{a}^{3}}=512 \\\ & \Rightarrow a=8 \\\ \end{aligned}$
Then our terms in G.P becomes $\dfrac{8}{r}$, $8$ and $8r$.
We are also given that when 4 is added to the first and second term, the three terms form an A.P. So, let us add 4 to the first and second term.
$\Rightarrow ~\dfrac{8}{r}+4$
$\Rightarrow 8+4=12$
So, our terms become $\dfrac{8}{r}+4$, $12$, $8r$. As we were given that they form an A.P, they must have a common difference.
$\Rightarrow 12-\left( \dfrac{8}{r}+4 \right)=8r-12$
Solving them we get,
$\begin{aligned} & \Rightarrow 12+12=8r+\left( \dfrac{8}{r}+4 \right) \\\ & \Rightarrow 8r+\dfrac{8}{r}+4=24 \\\ & \Rightarrow 8r+\dfrac{8}{r}=20 \\\ & \Rightarrow 8{{r}^{2}}+8=20r \\\ & \Rightarrow 8{{r}^{2}}-20r+8=0 \\\ & \Rightarrow 4\left( 2{{r}^{2}}-5r+2 \right)=0 \\\ & \Rightarrow \left( 2r-1 \right)\left( r-2 \right)=0 \\\ & \Rightarrow r=2,\dfrac{1}{2} \\\ \end{aligned}$
So, as we found the value of r, let us find the original G.P.
When $r=\dfrac{1}{2}$,
$\begin{aligned} & \Rightarrow \dfrac{8}{r}=\dfrac{8}{\dfrac{1}{2}}=8\times 2=16 \\\ & \Rightarrow 8r=8\times \dfrac{1}{2}=4 \\\ \end{aligned}$
Then the G.P is 16, 8, 4.
When $r=2$,
$\begin{aligned} & \Rightarrow \dfrac{8}{r}=\dfrac{8}{2}=4 \\\ & \Rightarrow 8r=8\times 2=16 \\\ \end{aligned}$
Then the G.P is 4, 8, 16.
As we take a look at both G.P they are in reverse order but the numbers are the same.
As we need to find the sum of terms of G.P,
$\begin{aligned} & \Rightarrow 4+8+16 \\\ & \Rightarrow 28 \\\ \end{aligned}$
Sum of terms of original G.P is 28.
Hence answer is Option D.

Note: There is a possibility of one making a mistake by confusing between A.P, G.P and H.P. one need to remember that three terms a, b, c are said to be in

1. A.P if they have a common difference, that is $b-a=c-b$.
2. G.P if they have common ratio, that is $\dfrac{b}{a}=\dfrac{c}{b}$.
3. H.P if their inverses are in A.P, that is their inverses have a common difference, that is $\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{1}{c}-\dfrac{1}{b}$.