Question

Find four numbers in G.P.such that their
product is 1 and the sum of the middle two
numbers is 10/3

Answer 1

$\huge\mathfrak\green{ }$

given series is 0.7 + 0.77 + 0.777 + ......... to nth term

therefore ,

$S_n$= 7×(0.1 + 0.11 + 0.111 +......to nth term)

=7/9 ( 0.9 + 0.99 + 0.999 + ...... to nth term )

=7/9 ( 9/10 + 99/100 + 999/1000 + ..... to nth term )

= 7/9 { ( 1 - 1/10 ) + ( 1 - 1/100 ) + ( 1 - 1/1000 ) + .... to nth term }

= 7/9 { 1+1+1+....to nth term } - { 1/10 + 1/100 + 1/1000 +.... to nth term}

= 7/9 { n - 1/10 ( 1 - 1/10 ) ^n / 1 - 1/10 }

= 7/9 { n - 1 ( 1 - 1/10 )^n / 9 }

= 7/9 { n - ( 1 - 1/10 )^n / 9 }

$\huge\mathfrak\blue{ proved }$

Answer 2

Step-by-step explanation:

Let the 4 numbers are

\begin{gathered}\frac{a}{ {r}^{3} } ,\: \: \frac{a}{r} ,\: \: ar ,\: \: a{r}^{3} \\\end{gathered}r3a,ra,ar,ar3

now according to the question,there product is 1:

\begin{gathered}\frac{a}{ {r}^{3} } \times \frac{a}{r} \times ar \times a{r}^{3} = 1\\ \\ {a}^{4} = 1 \\ \\ a = + - 1 \\ \\\end{gathered}r3a×ra×ar×ar3=1a4=1a=+−1

Sum of middle two numbers is 10/4

\begin{gathered}\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} \\ \\\end{gathered}ra+ar=310puta=1r1+r=3103(1+r2)=10r3r2−10r+3=03r2−9r−r+3=03r(r−3)−1(r−3)=0(3r−1)(r−3)=0r=3orr=31

So 4 numbers are

\begin{gathered}\frac{1}{27 } ,\: \: \frac{1}{3}, \: \: 3 ,\: \: 27 \\ \\ 27 ,\: 3, \: \frac{1}{3} ,\: \frac{1}{27} \\ \\\end{gathered}271,31,3,2727,3,31,271

and many more possible numbers also exist.

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