Question

Find four numbers in G.P. whose sum is 85 and product is 4096.

Answer 1

Hint: Let the four numbers in G.P. be $\dfrac{a}{{{r}^{3}}},\dfrac{a}{r},ar,a{{r}^{3}}$. Multiply these and equate with 4096 to get a = 8. Now, take their sum to get $\dfrac{8}{{{r}^{3}}}+\dfrac{8}{r}+8r+8{{r}^{3}}=85$. Next, let $z=r+\dfrac{1}{r}$. Substitute this in the previous equation to get $8{{z}^{3}}-16z-85=0$. Now, substitute \[t=2z\] to get ${{t}^{3}}-8t-85=0$. Solve for t and then find the value of r. Using values of r and a, find the four numbers.

Complete step-by-step answer:

In this question, we need to find four numbers that are in G.P. whose sum is 85 and product is 4096.

Let the four numbers in G.P. be $\dfrac{a}{{{r}^{3}}},\dfrac{a}{r},ar,a{{r}^{3}}$.

We are given that the product of these four numbers is equal to 4096.

So, ${{a}^{4}}=4096$

Taking fourth root of the above equation, we will get the following:

a = 8

So, the four numbers are $\dfrac{8}{{{r}^{3}}},\dfrac{8}{r},8r,8{{r}^{3}}$

Now, we are given that the sum of these four numbers is equal to 85.

$\dfrac{8}{{{r}^{3}}}+\dfrac{8}{r}+8r+8{{r}^{3}}=85$

Taking 8 common in the LHS, we will get the following:

$8\left( \dfrac{1}{{{r}^{3}}}+\dfrac{1}{r}+r+{{r}^{3}} \right)=85$

On rearranging the terms, we will get the following:

$8\left( {{r}^{3}}+\dfrac{1}{{{r}^{3}}} \right)+8\left( r+\dfrac{1}{r} \right)-85=0$
                              …(1)
Now, let $z=r+\dfrac{1}{r}$

${{z}^{3}}={{\left( r+\dfrac{1}{r} \right)}^{3}}$

${{z}^{3}}={{r}^{3}}+\dfrac{1}{{{r}^{3}}}+3r\cdot \dfrac{1}{r}\left( r+\dfrac{1}{r} \right)$

${{z}^{3}}={{r}^{3}}+\dfrac{1}{{{r}^{3}}}+3z$

So, ${{r}^{3}}+\dfrac{1}{{{r}^{3}}}={{z}^{3}}-3z$

So, equation (1) will become the following:

$8\left( {{z}^{3}}-3z \right)+8z-85=0$

$8{{z}^{3}}-16z-85=0$

Now, we will substitute \[t=2z\].

On doing this, we will get the following:

${{t}^{3}}-8t-85=0$

On factoring the above expression, we will get the following:

$\left( t-5 \right)\left( {{t}^{2}}+5t+17 \right)=0$

In this equation, $\left( {{t}^{2}}+5t+17 \right)=0$ gives imaginary roots. So, we will neglect those.

So, t = 2z = 5

$2\left( r+\dfrac{1}{r} \right)=5$
                                   …(2)
On simplifying this, we will get the following:

$2{{r}^{2}}-5r+2=0$

On factoring the above expression, we will get the following:

\[\left( r-2 \right)\left( 2r-1 \right)=0\]

So, \[r=2,\dfrac{1}{2}\] and \[a=8\].

Hence, the four numbers are 1, 4, 16, 64 or 64, 16, 4, 1.

Note: In this question, it is very important to note that the four numbers in G.P. to be $\dfrac{a}{{{r}^{3}}},\dfrac{a}{r},ar,a{{r}^{3}}$ and not $a,ar,a{{r}^{2}},a{{r}^{3}}$ or some other combination because it makes the calculation easier when we take the product of these. In the product, r is eliminated, and we are left only with a. so, whenever such a question comes, always choose the numbers in this way.