Question

Find three numbers in G.P such that their sum is 21, and the sum of their squares is 189​

Answer 1

The three numbers are in G.P. "(3, 6, 12) or (12, 6 , 3)".

Step-by-step explanation:

Let the required number be $\frac{a}{r}$, a and ar.

∴  $\frac{a}{r}$ + a + ar = 21   ......(1)

and

$\frac{a^{2} }{r^{2} }$ + $a^{2}$ + $a^{2}$$r^{2}$ = 91 ... (2)

On squaring both sides of (1), we get

$([tex]\frac{a}{r}$ + a + ar)^{2}[/tex] = 441

⇒$\frac{a^{2} }{r^{2} }$ + $a^{2}$ + $a^{2}$$r^{2}$ + 2a($\frac{a}{r}$ + a + ar) = 441

⇒ 189 + 2a(21) = 441

⇒  42a = 252

⇒  a = 6

Put a = 6 in (1), we get

6$r^{2}$ - 15r + 6 = 0

⇒6$r^{2}$ - 12r - 3r + 6 = 0

⇒ (r - 2)((6r - 3) = 0

∴ r = 2 or $\frac{1}{2}$

Hence, the three numbers are in G.P. "(3, 6, 12) or (12, 6 , 3)".