The sum of the first three terms of a GP 13/12 is and their product is - 1. Find the common ratio Class 9

The sum of the first three terms of a GP 13/12 is and their product is - 1. Find the common ratio Class 9

Question 1.

The sum of the first three terms of a GP $\frac{13}{12}$ is and their product is - 1. Find the common ratio and the terms. Class 9

Solution:

Let the first three terms of GP be a, ar, ar²

Accroding to the given conditions, a + ar + ar² = $\frac{13}{12}$ ...(i)

and a × ar × ar² = -1

∴ a³r³ = -1

∴ (ar)³ = (-1)³

∴ ar = -1

∴ a = $\frac{-1}{r}$

Substitute a = $\frac{-1}{r}$ in (i),

$\frac{-1}{r}+\left(\frac{-1}{r}\right) r+\left(\frac{-1}{r}\right) r^2=\frac{13}{12}$

Multiplying equation by 12r, we get

∴ -12 - 12r - 12r² = 13r

∴ 12r² + 25r + 12 = 0

∴ 12r² + 16r + 9r + 12 = 0

∴ 4r(3r + 4) + 3 (3r + 4) = 0

∴ (4r + 3) (3r + 4) = 0

∴ 4r + 3 = 0 or 3r + 4 = 0

∴ 4r = -3 or 3r = -4

∴ r = [latex]\frac{-3}{4}[/latex] or r = [latex]\frac{-4}{3}[/latex]

Case 1: For r = [latex]\frac{-3}{4}[/latex]

[latex]a=\frac{-1}{r}=\frac{-1}{\frac{-3}{4}}=\frac{4}{3}[/latex]

∴ The first three term of GP are

$\begin{aligned} & \frac{4}{3}, \frac{4}{3}\left(\frac{-3}{4}\right), \frac{4}{3}\left(\frac{-3}{4}\right)^2 \\ & \text { i.e., } \frac{4}{3},-1, \frac{3}{4}\end{aligned}$

Case 2: For r = $\frac{-4}{3}$

$a=\frac{-1}{r}=\frac{-1}{\frac{-4}{3}}=\frac{3}{4}$

∴ The first three term of GP are

$\begin{aligned} & \frac{3}{4^{\prime}}, \frac{3}{4}\left(\frac{-4}{3}\right), \frac{3}{4}\left(\frac{-4}{3}\right)^2 \\ & \text { i.e., } \frac{3}{4^{\prime}}-1, \frac{4}{3}\end{aligned}$

∴ The common ratio is $\frac{-3}{4}$ and the three terms are $\frac{4}{3^{\prime}}-1, \frac{3}{4}$

OR

∴ The common ratio is $\frac{-4}{3}$ and the three terms are $\frac{3}{4^{\prime}}-1, \frac{4}{3}$