Question

If x - 2, x+1 and x+7 are first three terms of a gp then find the value of x and find 11 term of it​

Answer 1

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Answer 2

The value of x is 5 and the first 11 terms are 3,6,12,24,48,96,192,384,768,1536,3072

Step-by-step explanation:

x - 2, x+1 and x+7 are first three terms of a gp

So, r = $\frac{a_2}{a_1}=\frac{a_3}{a_2}$

So, r =$\frac{x+1}{x-2}=\frac{x+7}{x+1}$

$\Rightarrow = (x+1)(x+1)=(x+7)(x-2)\\(x+1)^2=x^2-2x+7x-14\\x^2+1+2x=x^2+5x-14$

1+2x=5x-14

15=3x

x=5

So, Terms are :

x-2=5-2=3

x+1=5+1=6

x+7=5+7=12

So, r = $\frac{6}{3}=\frac{12}{6}=2$

a = first term = 3

r = 2

Formula of nth term :

$A_n=ar^{n-1}$

$A_4=3(2^{4-1})=24\\A_5=3(2^{5-1})=48\\A_6=3(2^{6-1})=96\\A_7=3(2^{7-1})=192\\A_8=3(2^{8-1})=384\\A_9=3(2^{9-1})=768\\A_{10}=3(2^{10-1})=1536\\A_{11}=3(2^{11-1})=3072$

Hence The value of x is 5 and the first 11 terms are 3,6,12,24,48,96,192,384,768,1536,3072

#Learn more:

Find the value of x such that x, x + 3 comma X + 9 are three terms of a GP

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