Question

If for an A.P. ${{a}_{1}},{{a}_{2}},{{a}_{3}},.......,{{a}_{n}},....$, ${{a}_{1}}+{{a}_{3}}+{{a}_{5}}=-12$ and ${{a}_{1}}{{a}_{2}}{{a}_{3}}=8$ then the value of ${{a}_{2}}+{{a}_{4}}+{{a}_{6}}$ equals to
A. $-12$
B. $-16$
C. $-18$
D. $-21$

Answer 1

Hint: We are given the arithmetic progression sequence, we will apply the conventional formula for ${{n}^{th}}$ term given by ${{a}_{n}}=a+\left( n-1 \right)d$ and find the first three terms for the series given in the question. Then we will find two equations in terms of a and d and then we will find their values by factorizing the obtained equations. Then we will find the terms required for the equation for which we have to find the value.

Complete step by step answer:
Since it is given that the sequence ${{a}_{1}},{{a}_{2}},{{a}_{3}},.......,{{a}_{n}},....$ is in arithmetic progression, therefore it will be in the following form:
$a,a+d,a+2d,.......,a+\left( n-1 \right)d,.....$
Where, ${{a}_{n}}=a+\left( n-1 \right)d$ is the ${{n}^{th}}$ term and $a$ is the first term in the sequence and $d$ is the common difference between terms.
Let’s take the first equation given in the equation that is: ${{a}_{1}}+{{a}_{3}}+{{a}_{5}}=-12\text{ }.........\text{Equation 1}\text{.}$ We will first find the values of these terms in terms of $a$ and $d$ :
We know that: ${{a}_{n}}=a+\left( n-1 \right)d$ .
Therefore,
$\begin{align}
  & {{a}_{1}}=a+\left( 1-1 \right)d\Rightarrow {{a}_{1}}=a \\
 & {{a}_{3}}=a+\left( 3-1 \right)d\Rightarrow {{a}_{3}}=a+2d \\
 & {{a}_{5}}=a+\left( 4-1 \right)d\Rightarrow {{a}_{5}}=a+4d \\
\end{align}$
Now we will put these values in equation 1:
 $\begin{align}
  & {{a}_{1}}+{{a}_{3}}+{{a}_{5}}=-12 \\
 & \Rightarrow a+\left( a+2d \right)+\left( a+4d \right)=-12 \\
 & \Rightarrow 3a+6d=-12\Rightarrow 3\left( a+2d \right)=-12 \\
 & \Rightarrow a+2d=-4\text{ }...........\text{ Equation 2}\text{.} \\
\end{align}$
We will now take the second condition given in the question that is: ${{a}_{1}}{{a}_{2}}{{a}_{3}}=8\text{ }.........\text{ Equation 3}\text{.}$ We already found the values of ${{a}_{1}}\text{=}a$ and ${{a}_{3}}=a+2d$. Now we will find the value of ${{a}_{2}}$
${{a}_{2}}=a+\left( 2-1 \right)d\Rightarrow {{a}_{2}}=a+d$
Putting these values in equation 3,
 $\begin{align}
  & {{a}_{1}}{{a}_{2}}{{a}_{3}}=8\text{ } \\
 & \Rightarrow \left( a \right)\left( a+d \right)\left( a+2d \right)=8\text{ }........\text{Equation 4}\text{.} \\
\end{align}$
Now from equation 2 we have: $a=-4-2d$ , we will put it in equation 4:
\[\begin{align}
  & \Rightarrow \left( a \right)\left( a+d \right)\left( a+2d \right)=8\text{ } \\
 & \Rightarrow \left( -4-2d \right)\left( -4-2d+d \right)\left( -4-2d+2d \right)=8 \\
 & \Rightarrow \left( -4-2d \right)\left( -4-d \right)\left( -4 \right)=8\Rightarrow \left( -4-2d \right)\left( -4-d \right)=-2 \\
 & \Rightarrow \left( 4+2d \right)\left( 4+d \right)=-2\Rightarrow 16+4d+8d+2{{d}^{2}}+2=0 \\
 & \Rightarrow {{d}^{2}}+6d+9=0 \\
 & \Rightarrow {{d}^{2}}+3d+3d+9=0\Rightarrow \left( d+3 \right)\left( d+3 \right)=0 \\
 & \Rightarrow d=-3 \\
\end{align}\]
From equation 2 we have: $a=-4-2d$, putting the value of d here , we will get : $a=-4-(2\times -3)=2\Rightarrow a=2$
Now we have to find the value of: ${{a}_{2}}+{{a}_{4}}+{{a}_{6}}$ ,
$\begin{align}
  & {{a}_{2}}=a+\left( 2-1 \right)d\Rightarrow {{a}_{2}}=a+d \\
 & {{a}_{4}}=a+\left( 4-1 \right)d\Rightarrow {{a}_{4}}=a+3d \\
 & {{a}_{6}}=a+\left( 6-1 \right)d\Rightarrow {{a}_{6}}=a+5d \\
\end{align}$
Putting these into the required equation:
${{a}_{2}}+{{a}_{4}}+{{a}_{6}}=\left( a+d \right)+\left( a+3d \right)+\left( a+5d \right)\Rightarrow 3a+9d$
After putting the values of a and d we will get:
$\begin{align}
  & 3a+9d=\left( 3\times 2 \right)+\left( 9\times -3 \right)=6-27 \\
 & \Rightarrow -21 \\
\end{align}$

So, the correct answer is “Option D”.

Note: You need not show the derivation of each term again and again you can also write it directly. It is written here for your understanding. Take care while solving the equations, since there is a lot of sign changing taking place, the chances of making mistakes are more there.