The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Class 9

The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Class 9

Question 1.

The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP. Class 9

Solution:

Let the first three terms of GP be

a, ar, ar²

According to the given conditions,

a + ar + ar² = 26

∴ a(1 + r + r²) = 26 ...(i)

and a² + (ar)² + (ar²)² = 364

∴ a² + a²r² + a²r⁴ = 364

∴ a²(1 + r² + r⁴) = 364 ...(ii)

Dividing (ii) by (i), we get

$\begin{array}{ll} & \frac{a^2\left(1+r^2+r^4\right)}{a\left(1+r+r^2\right)}=\frac{364}{26} \\ \therefore & a \times \frac{\left(1+r^2+r^4\right)}{\left(1+r+r^2\right)}=14 \\ \therefore & a \times \frac{\left(1+r+r^2\right)\left(1-r+r^2\right)}{\left(1+r+r^2\right)}=14\end{array}$

... [∵ 1 + r² + r⁴ = (1 + r + r²)(1 - r + r²)]

∴ a(1 - r + r²) = 14 ... (iii)

Dividing (i) by (iii), we get

$\begin{aligned} & \frac{a\left(1+r+r^2\right)}{a\left(1-r+r^2\right)}=\frac{26}{14} \\ & \frac{1+r+r^2}{1-r+r^2}=\frac{13}{7}\end{aligned}$

∴ 7 (1 + r + r²) = 13(1 - r + r²)

∴ 7 + 7r + 7r² = 13 - 13r + 13r²

∴ 6r² - 20r + 6 = 0

Dividing both side by 2, we get

3r² - 10r + 3 = 0

∴ 3r² - 9r - r + 3 = 0

∴ 3r (r - 3) - 1 (r - 3) = 0

∴ (3r - 1) (r - 3) = 0

∴ 3r - 1 = 0 or r - 3 = 0

∴ r = - or r = 3

Case 1: For r = $\frac{1}{3}$

$a\left(1+\frac{1}{3}+\frac{1}{9}\right)=26 \quad \ldots[$ From (i) $]$

$\begin{aligned} & a \times \frac{13}{9}=26 \\ & a=26 \times \frac{9}{13}=18\end{aligned}$

∴ The first three terms of GP are

18, $18\left(\frac{1}{3}\right), 18\left(\frac{1}{3}\right)^2$

i.e., 18, 6, 2

Case 2: For r = 3

a (1 + 3+ 9) = 26 ....[From (i)]

13a = 26

a = 2

∴ The first three terms of GP are

2, 2(3), 2(3)²

i.e., 2, 6, 18