Question

Let \[{{a}_{1}},{{a}_{2}},{{a}_{3}},...\] be an A.P. with \[{{a}_{6}}=2\]. Then, what will be the common difference of this A.P., which maximizes the product \[{{a}_{1}}{{a}_{4}}{{a}_{5}}\]?
A. \[\dfrac{6}{5}\]
B. \[\dfrac{8}{5}\]
C. \[\dfrac{2}{3}\]
D. \[\dfrac{3}{2}\]

Answer 1

Hint: An AP is a series in which the first term is a and the terms have a common difference between them, d. We will start by finding the term \[{{a}_{6}}=2\] using the formula for ${{n}^{th}}$ term of an AP given by ${{a}_{n}}=a+\left( n-1 \right)d$ . So, we will get a in terms of d. Again, find the terms, \[{{a}_{1}},{{a}_{4}},{{a}_{5}}\] using the formula and then compute the product by substituting a in terms of d from the above results. Then, to find the d which maximizes the product \[{{a}_{1}}{{a}_{4}}{{a}_{5}}\], we have to use the second derivative test. So, we will compute f’(d) first, equate to 0 and get values of d. Then find f”(d) at obtained values of d. The value of d at which we get f”(d) < 0 , will give us the answer.

Complete step by step answer:
We have been given that \[{{a}_{1}},{{a}_{2}},{{a}_{3}},...\] are in A.P and \[{{a}_{6}}=2\] .
Let us consider a is the first term and d is a common difference of the AP. Then we know that the formula for ${{n}^{th}}$ term of an AP is given by ${{a}_{n}}=a+\left( n-1 \right)d$ .
Now, according to given condition, \[{{a}_{6}}=2\], we get
\[\begin{align}
  & {{a}_{6}}=a+\left( 6-1 \right)d \\
 & 2=a+5d \\
 & a+5d=2 \\
 & a=2-5d.............(i) \\
\end{align}\]
Now, we have to find the product \[{{a}_{1}}{{a}_{4}}{{a}_{5}}\] so we can use the formula as below,

\[\begin{align}
  & {{a}_{1}}=a \\
 & {{a}_{4}}=a+\left( 4-1 \right)d\Rightarrow a+3d \\
 & {{a}_{5}}=a+\left( 5-1 \right)d\Rightarrow a+4d \\
 & \therefore {{a}_{1}}{{a}_{4}}{{a}_{5}}=a(a+3d)(a+4d) \\
\end{align}\]
From equation (i), we have to substitute a and we have
\[\begin{align}
  & {{a}_{1}}{{a}_{4}}{{a}_{5}}=\left( 2-5d \right)\left( \left( 2-5d \right)+3d \right)\left( \left( 2-5d \right)+4d \right) \\
 & \Rightarrow {{a}_{1}}{{a}_{4}}{{a}_{5}}=\left( 2-5d \right)\left( 2-5d+3d \right)\left( 2-5d+4d \right) \\
 & \Rightarrow {{a}_{1}}{{a}_{4}}{{a}_{5}}=\left( 2-5d \right)\left( 2-2d \right)\left( 2-d \right) \\
\end{align}\]
On simplifications, we get
\[\begin{align}
  & \Rightarrow {{a}_{1}}{{a}_{4}}{{a}_{5}}=-\left( 5d-2 \right)\left( 2d-2 \right)\left( d-2 \right) \\
 & \Rightarrow {{a}_{1}}{{a}_{4}}{{a}_{5}}=-2\left( \left( 5d-2 \right)\left( d-1 \right)\left( d-2 \right) \right) \\
\end{align}\]
Now, if we expand the terms by opening the brackets, the equation becomes
\[\begin{align}
  & \Rightarrow {{a}_{1}}{{a}_{4}}{{a}_{5}}=-2\left( \left( 5d-2 \right)\left( {{d}^{2}}-3d+2 \right) \right) \\
 & \Rightarrow {{a}_{1}}{{a}_{4}}{{a}_{5}}=-2\left( 5{{d}^{3}}-15{{d}^{2}}+10d-2{{d}^{2}}+6d-4 \right) \\
 & \Rightarrow {{a}_{1}}{{a}_{4}}{{a}_{5}}=-2\left( 5{{d}^{3}}-17{{d}^{2}}+16d-4 \right) \\
\end{align}\]
We have got the product. Now, we have been asked to find the maximum value. We know that to find the maximum or minimum values of a function, we make use of the second derivative test. So, first we will compute the first derivative, equate it to 0 and get the critical point. Then we will find the second derivative and find the value at that critical point. If it is > 0, then we have a minima and if it is < 0 we have a maxima condition.
Let us consider \[f\left( d \right)=-2\left( 5{{d}^{3}}-17{{d}^{2}}+16d-4 \right)\]
So, on differentiating the above equation, we get
\[\begin{align}
  & f'\left( d \right)=-2\left( 3\times 5{{d}^{2}}-2\times 17d+16\times 1+0 \right) \\
 & f'\left( d \right)=-2\left( 15{{d}^{2}}-34d+16 \right) \\
\end{align}\]
We can rewrite this equation as,
\[\begin{align}
  & f'\left( d \right)=-2\left( 15{{d}^{2}}-10d-24d+16 \right) \\
 & f'\left( d \right)=-2\left( 5d\left( 3d-2 \right)-8\left( 3d-2 \right) \right) \\
 & f'\left( d \right)=-2\left( 5d-8 \right)\left( 3d-2 \right) \\
\end{align}\]
Equating it to 0, we have
\[\begin{align}
  & \Rightarrow f'\left( d \right)=0 \\
 & \Rightarrow 5d-8=0\text{ or }3d-2=0 \\
\end{align}\]
\[\Rightarrow d=\dfrac{8}{5}\text{ or }d=\dfrac{2}{3}\]
Now, let us find second derivative as below,
\[\begin{align}
  & f'\left( d \right)=-2\left( 15{{d}^{2}}-34d+16 \right) \\
 & f''\left( d \right)=-2\left( 2\times 15d-34\times 1+0 \right) \\
 & f''\left( d \right)=-2\left( 30d-34 \right) \\
\end{align}\]
Let us check the value of f’’(d) at \[d=\dfrac{8}{5}\] . So, we have
\[\begin{align}
  & f''\left( \dfrac{8}{5} \right)=-2\left( 30\times \dfrac{8}{5}-34 \right) \\
 & \Rightarrow -2\left( 48-34 \right) \\
 & \Rightarrow -2\times 14 \\
 & \Rightarrow -28 \\
\end{align}\]
Since, we have got f”(d) < 0 , we get that at \[d=\dfrac{8}{5}\] , the product is maximum.
Now let us check value of f’’(d) at \[d=\dfrac{2}{3}\]
\[\begin{align}
  & f''\left( \dfrac{2}{3} \right)=-2\left( 30\times \dfrac{2}{3}-34 \right) \\
 & \Rightarrow -2\left( 20-34 \right) \\
 & \Rightarrow -2\times -14 \\
 & \Rightarrow 28 \\
\end{align}\]
Since, we have got f”(d) > 0 , we get that at \[d=\dfrac{2}{3}\] , the product is minimum.
Thus, from the above results, we can conclude that the the common difference of this A. P., d, which maximizes the produce \[{{a}_{1}}{{a}_{4}}{{a}_{5}}\] is \[d=\dfrac{8}{5}\].

So, the correct answer is “Option B”.

Note: Students can make mistakes while solving derivation. They can forget to check the second derivative \[f''(d)\]. As there is an option \[\dfrac{2}{3}\] in the question, one can select this option as the correct one. But they must note that option (C) \[\dfrac{2}{3}\] is incorrect. After solving the second derivative only, they can see the value of \[f''(d)\] is less than 0 for \[\dfrac{8}{5}\]. Calculation plays an important role in these types of problems, so do not miss any steps and write them directly as there is a chance of errors.