Question

Find the numbers in G.P a/r, a , ar , such that there sum is 42 and there product is 1728.​

Answer 1

Answer:

6, 12 and 24 or 24, 12, 6

Step-by-step explanation:

Let the numbers are a/r, a, ar, where r is common ratio.

Their product = 1728

=> (a/r) * a * (ar) = 1728

=> a³ = 1728

=> a³ = 12³

=> a = 12

Also, their sum = 42

=> a/r + a + ar = 42

=> 12/r + 12 + 12r = 42

=> 12/r + 12r - 30 = 0

=> (12 + 12r² - 30r) = 0

=> 6(2 + 2r² - 5r) = 0

=> 2r² - 5r + 2 = 0

=> (r - 2)(2r - 1) = 0

=> r = 2 or 1/2

Hence, terms are:

a/r = 12/2 or 12/(1/2) → 6 or 24

a = 12

ar = 12(2) or 12(1/2) → 24 or 6

Answer 2

G. P.

$\frac{a}{r} \: a \: ar$

$Common ratio, r = \frac{a}{ \frac{a}{r} } = r$

Sum of G. P.

$S_n = \frac{a( {r}^{n} - 1) }{r - 1} \\ 42 = \frac{ \frac{a}{r}( {r}^{n} - 1) }{r - 1} \: \: \: \: \: \: \: \: ...(1)$

Also, product of G. P. is 1728

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

Now, putting a in (1)

$42 = \frac{ \frac{12}{r}( {r}^{3} - 1) }{r - 1} \\ 42(r - 1) = \frac{12}{r} ( {r}^{3} - 1) \\ 42 {r}^{2} - 42r = 12 {r}^{3} - 12 \\ 42 ({r}^{2} - r )= 12( {r}^{3} - 1) \\ 7 {r}^{2} - 7r = 2 {r}^{3} - 2 \\ 2{r}^{3} - 7 {r}^{2} + 7r - 2 = 0$

Solving this we get

$r = 1 \\ r = \frac{1}{2 } \\ r = 2$

When r=1

G. P. is 12,12,12

When r=1/2

G. P. is 24,12,6

When r=2

G.P. is 6,12,24