Question

The numbers 3, x and x + 6 form are in G.P.
Find (i), (ii) 20th term (iii) nth term.

Answer 1

a b & c are in gp then

b^2 = ac


X^2 = 3 ( x + 6)


X^2 - 3 x - 18 = 0

X = 6

X = - 3 invalid

Then no. are


3 6 12 24 48..............

Ur responsibility to solve Rest part

Answer 2

Answer: Hence, ii) $a_{20}=1572864$, $a_{20}=-3$ (iii) $a_n=3(2)^{n-1}$,$a_n=3(-1)^{n-1}$

Step-by-step explanation:

Since we have given that

3, x and x+6.

Since they are in G.P.so, we know that

$x^2=3(x+6)\\\\x^2=3x+18\\\\x^2-3x-18=0\\\\x^2-6x+3x-18=0\\\\x(x-6)+3(x-6)=0\\\\x=6,-3$

If x = 6, then terms would be 3,6,12.

So, Value of 20 th term would be

$a_{20}=ar^{n-1}=3\times (2)^{20-1}=3\times 20^{19}=1572864$

And the nth value would be

$a_n=3(2)^{n-1}$

If x = -3, then terms would be 3,-3, 3

$a_{20}=3(-1)^{20-1}=3(-1)^{19}=-3$

and the nth value would be

$a_n=3(-1)^{n-1}$

Hence, ii) $a_{20}=1572864$, $a_{20}=-3$ (iii) $a_n=3(2)^{n-1}$,$a_n=3(-1)^{n-1}$