Question

$\sf \large \:\underline{ \red{Given} }:-$
The sum of first three terms of a G.P. is $\frac{39}{10}$and their product is 1.
Find the common ratio and the terms. ​

Answer 1

Answer:

$\sf \huge \large \:\underline{ \red{Answer} }:-$

Let $\frac{a}{r}$, a, ar be the first three terms of the G.P.

$\huge{ \frac{a}{r} + (a)(ar) = 1...(1)}$

From (2), we obtain

$\huge{ {a}^{2} = 1}$

=> a = 1 {considering real roots only }

Substituting a = 1 in equation (1), we obtain

$\huge{ \frac{1}{r} + 1 + r = \frac{39}{10} }$

$= > \huge{ 1 + r + {r}^{2} = \frac{39}{10} r }$

$= > {10r}^{2} \: 29r + 10 = 0$

$= > {10r}^{2} - {25}^{r} - {4}^{r} + 10 = 0$

$= > {10r}^{2} - {25}^{r} - 4r + 10 = 0$

$= > 5r \: (2r - 5) - 2(2r - 5) = 0$

$= > (5r - 2)(2r - 5 ) = 0$

$= > \huge{r = \frac{2}{5} \: or \: \frac{5}{2} }$

$\boxed { \sf \red{Thus,} the \: three \: terms \: of \: g.p \: are \: \frac{5}{2},1, and \frac{2}{5} }$

Answer 2

Step-by-step explanation:

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