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

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

& {{a}_{6}}=a+\left( 6-1 \right)d \\\ & 2=a+5d \\\ & a+5d=2 \\\ & a=2-5d.............(i) \\\ \end{aligned}$$ Now, we have to find the product $${{a}_{1}}{{a}_{4}}{{a}_{5}}$$ so we can use the formula as below, $$\begin{aligned} & {{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{aligned}$$ From equation (i), we have to substitute a and we have $$\begin{aligned} & {{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{aligned}$$ On simplifications, we get $$\begin{aligned} & \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{aligned}$$ Now, if we expand the terms by opening the brackets, the equation becomes $$\begin{aligned} & \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{aligned}$$ We have got the product. Now, we have been asked to find the maximum value. We know that to find the maximum or minimum values of a function, we make use of the second derivative test. So, first we will compute the first derivative, equate it to 0 and get the critical point. Then we will find the second derivative and find the value at that critical point. If it is > 0, then we have a minima and if it is < 0 we have a maxima condition. Let us consider $$f\left( d \right)=-2\left( 5{{d}^{3}}-17{{d}^{2}}+16d-4 \right)$$ So, on differentiating the above equation, we get $$\begin{aligned} & 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{aligned}$$ We can rewrite this equation as, $$\begin{aligned} & 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{aligned}$$ Equating it to 0, we have $$\begin{aligned} & \Rightarrow f'\left( d \right)=0 \\\ & \Rightarrow 5d-8=0\text{ or }3d-2=0 \\\ \end{aligned}$$ $$\Rightarrow d=\dfrac{8}{5}\text{ or }d=\dfrac{2}{3}$$ Now, let us find second derivative as below, $$\begin{aligned} & 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{aligned}$$ Let us check the value of f’’(d) at $$d=\dfrac{8}{5}$$ . So, we have $$\begin{aligned} & 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{aligned}$$ Since, we have got f”(d) < 0 , we get that at $$d=\dfrac{8}{5}$$ , the product is maximum. Now let us check value of f’’(d) at $$d=\dfrac{2}{3}$$ $$\begin{aligned} & 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{aligned}$$ Since, we have got f”(d) > 0 , we get that at $$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.