Question

find five numbers in GP such that their sum is 31/4 and product is 1​

Answer 1

Step-by-step Explanation

Given that:

Find five numbers in GP such that their sum is 31/4 and product is 1.

Solution: To find five numbers in GP,

it is always convenient to take those numbers

$\frac{a}{ {r}^{2} }, \: \frac{a}{r}, \: a, \: ar, \: a {r}^{2}$

Now according to the question

Sum of these are 31/4

$\frac{a}{ {r}^{2} } + \frac{a}{r} + a + ar + a {r}^{2} = \frac{31}{4}$

Product is 1

$\frac{a}{ {r}^{2} } \times \frac{a}{r} \times a \times ar \times a {r}^{2} = 1 \\ \\ {a}^{5} = 1 \\ \\ {a}^{5} = {1}^{5} \\ \\ a = 1$

Now put the value of a in eq 1

$\frac{1}{ {r}^{2} } + \frac{1}{r} + 1 + r + {r}^{2} = \frac{31}{4}$

Now manipulate the terms,

$(r + \frac{1}{r} ) + ( {r}^{2} + \frac{1}{ {r}^{2} } ) + 1 = \frac{31}{4} \\ \\ (r + \frac{1}{r} ) + ( {r}^{2} + \frac{1}{ {r}^{2} } ) + 2 - 2 + 1 = \frac{31}{4} \\ \\ {(r + \frac{1}{r} })^{2} + (r + \frac{1}{r} ) - 1 - \frac{31}{4} = 0 \\ \\ {(r + \frac{1}{r} })^{2} + (r + \frac{1}{r} ) - \frac{35}{4} = 0$

It is a quadratic equation in (r+1/r),

For simplification put (r+1/r)= t

${t}^{2} + t - \frac{35}{4} = 0 \\ \\ here \: a = 1 \\ \\ b = 1 \\ \\ c = \frac{ - 35}{4}$

Apply quadratic formula

$t_{1,2} = \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\t_{1,2} = \frac{ - 1 ± \sqrt{ {1}^{2} - 4 \times 1 \times \frac{ - 35}{4} } }{2} \\ \\ t_{1,2} = \frac{ - 1 ± \sqrt{1 + 35} }{2} \\ \\ t_{1,2} = \frac{ - 1 ± \sqrt{36} }{2} \\ \\ t_1 = \frac{ - 1 + 6}{2} \\ \\ t_1= \frac{ 5}{2} \\ \\ t_2 = \frac{ - 7}{2}$

Put the value of t1 and t2

$r + \frac{1}{r} = \frac{5}{2} \\ \\ 2({r}^{2} + 1) = 5r \\ \\ 2 {r}^{2} - 5r + 2 = 0 \\ \\ 2 {r}^{2} - 4r - r + 2 = 0 \\ \\ 2r(r - 2) - 1(r - 2) = 0 \\ \\ (r - 2)(2r - 1) = 0 \\ \\ \bold{\red{r = 2}} \\ \\\bold{\blue{ r = \frac{ 1}{2}}}$

By putting another value of t

$r + \frac{1}{r} = \frac{ - 7}{2} \\ \\ 2( {r}^{2} + 1) = - 7r \\ \\ 2 {r}^{2} + 7r + 2 = 0 \\ \\ r_{1,2} = \frac{ - 7 ± \sqrt{49 - 16} }{4} \\ \\ r_{1,2} = \frac{ - 7 ± \sqrt{33} }{4}$

Thus the terms of GP are

$\bold{\frac{1}{4} ,\: \frac{1}{2} , \: 1 ,\: 2, \: 4}$

Another

$\bold{\frac{16}{( - 7 + \sqrt{33} )^{2} } , \: \frac{4}{ - 7 + \sqrt{33} } , \: 1, \: \frac{- 7 + \sqrt{33}}{4} , \:\frac{ ( - 7 + \sqrt{33} )^{2}}{16} }$

and

$\bold{\frac{16}{( - 7 - \sqrt{33} )^{2} } , \: \frac{4}{ - 7 - \sqrt{33} } , \: 1, \: \frac{- 7 - \sqrt{33}}{4} , \:\frac{ ( - 7 - \sqrt{33} )^{2}}{16} }$

Hope it helps you.