Question

find four numbers in GP such that sum of middle two number is 10/3 and their product is 1​

Answer 1

Answer:

$27,3,\frac{1}{3},\frac{1}{27}\ \ \ \ or\ \ \ \frac{1}{27},\frac{1}{3},3,27$

Step-by-step explanation:

As per the question,

Let the number in G.P are $\frac{a}{r^{3}}\ ,\ \frac{a}{r}\ ,\ ar\ ,\ ar^{3}$

According to the question,

Sum of the middle two term = 10/3

That is,

$\frac{a}{r}+ar=\frac{10}{3}$

And

Product = 1

$\frac{a}{r}\times ar=1\\a^{2} = 1$

∴ a = 1

Put the value of a in

$\frac{a}{r}+ar=\frac{10}{3}$

$\frac{1}{r}+1r=\frac{10}{3}$

$3r^{2}-10r+3=0$

$(3r-1)(r-3)=0$

∴ r = 1/3 or r = 3

If r = 1/3

$\frac{a}{r^{3}} = 27\\\frac{a}{r} = 3\\ar =\frac{1}{3}\\ar^{3} = \frac{1}{27}$

If r = 3

$\frac{a}{r^{3}} = \frac{1}{27}\\\frac{a}{r} = \frac{1}{3}\\ar =3\\ar^{3} = 27$

Therefore,

The required four term of G.P = $27,3,\frac{1}{3},\frac{1}{27}\ \ \ \ or\ \ \ \frac{1}{27},\frac{1}{3},3,27$