Question

Find four numbers in G.P. such that their product is 1 and the sum of the middle two numbers is $\frac{10}{3}$

Answer 1

Let the 4 numbers are
$\frac{a}{ {r}^{3} } ,\: \: \frac{a}{r} ,\: \: ar ,\: \: a{r}^{3}$
now according to the question,there product is 1:

$\frac{a}{ {r}^{3} } \times \frac{a}{r} \times ar \times a{r}^{3} = 1\\ \\ {a}^{4} = 1 \\ \\ a = + - 1$
Sum of middle two numbers is 10/4

$\frac{a}{r} + ar = \frac{10}{3} \\ \\ put \: a = 1 \\ \\ \frac{1}{r} + r = \frac{10}{3} \\ \\ 3(1 + {r}^{2} ) = 10r \\ \\ 3 {r}^{2} - 10r + 3 = 0 \\ \\ 3 {r}^{2} - 9r - r+ 3 = 0 \\ \\ 3r(r - 3) - 1(r - 3) = 0 \\ \\ (3r - 1)(r - 3) = 0 \\ \\ r = 3 \\ \\ or \\ \\ r = \frac{1}{3}$
So 4 numbers are

$\frac{1}{27 } ,\: \: \frac{1}{3}, \: \: 3 ,\: \: 27 \\ \\ 27 ,\: 3, \: \frac{1}{3} ,\: \frac{1}{27}$
and many more possible numbers also exist.

Hope it helps you.

Answer 2

Refer to the attachment above for your answer!!!

Thank you!!...✔✔