Question

the three numbers are in G.P., and their sum and product are respectively 63 and 1728. Find
them
.​

Answer 1

Step-by-step explanation:

Let the three terms in GP be

$\frac{a}{r}$

$a$

and

$ar$

with common ratio 'r'.

Their product is 1728.

So,

$\frac{a}{r} \times a \times ar = 1728 \\ {a}^{3} = 1728 = {12}^{3} \\ a = 12$

So, the second term is 12.

So, the terms become 12/r, 12 and 12r respectively.

Now, Sum is 63.

$\frac{12}{r} + 12 + 12r = 63 \\ \frac{12}{r} + 12r = 51 \\ \frac{4}{r} + 4r = 17 \\ 4 + 4 {r}^{2} = 17r \\ 4 {r}^{2} - 17r + 4 = 0 \\ 4 {r}^{2} - 16r - r + 4 = 0 \\ 4r(r - 4) - 1(r - 4) = 0 \\ (4r - 1)(r - 4) = 0 \\ r = \frac{1}{4} \: or \: r = 4$

Case 1: If r=1/4, the terms are 12/(1/4), 12 and 12(1/4). They are 48, 12 and 3.

Case 2: If r=4, the terms are 12/4, 12 and 12(4). They are 3, 12 and 48.

Hence, in any order we can write.

The GP is 3, 12 and 48.

Thank you