Question

the numbers 3 ,X and X + 6 form are in a GP find (1)X,(2)20 term (3) n term​

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

Given that-

3, x and (x + 6) are in G.P.

Find-

1. x = ?
2. 20th term
3. n term

Solution:

1. The value of x

We know that if a, b and c are in G.P. then-

$b = \sqrt{ac}$

Here, a = 3, b = x and c =(x + 6)

So, $x = \sqrt{3(x+6)}$

              = $\sqrt{3x+ 18}$

Squaring on boths sides, we have-

$x^{2} = 3x + 18$

$x^{2} - 3x - 18 = 0$

$x^{2} -(6-3)x - 18 = 0$

$x^{2} - 6x + 3x - 18 = 0$

$x(x -6)+3(x-6) = 0$

$(x+3)(x-6) = 0$

x = -3, and 6

Neglecting the -ive sign.

We have x = 6  

Hence, the value of x is 6

2. The 20th term of G.P.

20th term of G.P. = ?

We have-

First term a = 3

Second term = x

                      = 6

Second term = 6

$Difference = \dfrac{Second \ term}{First \ term}$

$d = \dfrac{6}{3}$

r = 2

We know that

The nth term of G.P-

$T_{n} = a(r^{n-1})$

So, $T_{20} = 3.(2^{20-1)}$

                        = $3.(2^{19})$

$T_{20} = = 3 \times2^{19}$

                                      ANSWER

3. Find nth term-

We know that-

The nth term of any G.P.-

$T_{n} = a(r^{n-1})$