Question

find three numbers in GP such that their sum is 35 and their products is 1000​

Answer 1

Answer:

Let the Three numbers in GP be :-)

$\frac{a}{r}, a,ar$

Given that

product of these three e numbers is 1000.

so,

$\frac{a}{r} \times a \times ar = 1000 \\ \implies {a}^{3} = 1000 \\ \implies a = \sqrt[3]{10 \times 10 \times 10} \: \\ \implies \: \large \boxed { \boxed{a = 10}}$

also given that sum of three numbers is 35 .

So,

$\frac{a}{r} + a + ar = 35 \\ \implies \frac{10}{r} + 10 + 10r = 35 \\ \implies10 + 10r + 10 {r}^{2} = 35r \\ \implies10 {r}^{2} - 25r + 10 = 0 \\ \implies10 {r}^{2} - 20r - 5 r+ 10 = 0 \\ \implies10r(r - 2) - 5(r - 2) = 0 \\ \implies (10r - 5)(r - 2) = 0 \\ \implies \large \boxed{ \boxed{ r = 2 \: \: \: \: \: or \: \: \: \: r = \frac{1}{2} }}$

So ,

the required numbers using our assumption will be:-)

$5,10,20 \: \: \: \: \: or \: \: \: \: \: 20,10,5$

Answer 2

$\large \underline{ \underline{ \sf \: Solution : \: \: \: }}$

Let ,

Three numbers of G.P are $\sf \frac{a}{r},a,ar$

According to the question ,

Product of 3 numbers is 1000

$\to \sf a \times \frac{a}{r} \times ar\ =\ 1000 \\ \\ </p><p> \to \sf</p><p>a^{3}\ =\ 1000 \\ \\ </p><p> \to \sf</p><p>a=\ 10</p><p>$

Sum of three numbers is 35

$\to \sf</p><p>\frac{a}{r}\ +\ a\ +\ ar\ =\ 35</p><p> \\ \\ \to \sf</p><p>\frac{10}{r}\ +\ 10\ +\ 10r\ =\ 35 \\ \\ \to \sf</p><p>10r^{2}\ +\ 10r\ +\ 10\ =\ 35r</p><p> \\ \\ \to \sf</p><p>10r^{2}\ -\ 25r\ +\ 10\ =\ 0</p><p> \\ \\ \to \sf </p><p>2r^{2}\ -\ 5r\ +\ 2\ =\ 0$

On solving , we get

$\sf r = 2,\frac{1}{2}$

Therefore , the required numbers are 5 , 10 , 20