Question

the product of three consecutive terms of a gp is 343 and their sum is 91/3 . find the three terms
Pls answer asap ppl
Good morning btw​

Answer 1

The answer is here bro.

ANSWER!!!!!!!!!!!!!!!!!!

Let the three consecutive terms of gp be a/r,a,ar

Given that :

a/r.a.ar =343

a^3 = 343

a.a.a=7.7.7

a=7

a/r +a + ar =9/3

a + ar +ar^2 / r = 9/3

a(r^2 + r + 1)/r = 3

7(r^2 + r + 1)/r = 3

(r^2 + r + 1)/r = 3/7

(r^2 + r + 1) = 3r/7

7r^2 +7r + 7= 3r

7r^2 + 4r + 7= 0

the answer is 0

to abhi thanks button ko maar maar ke ura do.

Answer 2

Answer:

The terms are

7/3, 7 and 7.3

or

7.3, 7 and 7/3

Step-by-step explanation:

Given:

The product of 3 consecutive term of G.P = 343.

And there sum is = 91/3

To find:

The 3 terms

Solution:

Let, the 3 terms be x/a, x and a.x [here x is geometric mean and a is common multiple]

According to question,

$\frac{x}{a} \times x \times a.x = 343 \\ \\ x.x.x = 343 \\ \\ {x}^{3} = 343 \\ \\ x = 7$

Now, We have to find the value of a.

$\frac{x}{a} + x + a.x = \frac{91}{3} \\ \\ x( \frac{1}{a} + 1 + a) = \frac{91}{3} \\ \\ 7( \frac{1}{a} + a + 1) = \frac{91}{3} \\ \\ \frac{1}{a} + a + 1 = \frac{13}{3} \\ \\ \frac{1}{a} + a = \frac{13}{3} - 1 \\ \\ \frac{1}{a} + a = \frac{13 - 3}{3} \\ \\ \frac{1}{a} + a = \frac{10}{3} \\ \\ \frac{1 + {a}^{2} }{a} = \frac{10}{3} \\ \\ 3 + 3{a}^{2} = 10a \\ \\ 3 {a}^{2} - 10a + 3 = 0 \\ \\ 3 {a}^{2} - 9a - a + 3 = 0 \\ \\ 3a(a - 3) - 1(a - 3) = 0 \\ \\ (a - 3)(3a - 1) = 0 \\ \\ a - 3 = 0 \\ \\ a = 3 \\ \\ either \\ \\ 3a - 1 = 0 \\ \\ 3a = 1 \\ \\ a = \frac{1}{3}$

So, The common multiple is 3 or 1/3

Therefore, The terms are 7/3, 7 and 7.3 or 7.3, 7 and 7/3