Question

If the numbers 1\2x, x + 1 ,2 (x+3) are consecutive terms of a geometric sequence, what is the value of ? Find the value of those terms and the general term of the sequence in the simplest form.

Answer 1

Step-by-step explanation:

General form of geometric series is nth term is ar^(n-1). Here it can be seen that square

of middle term=product of n-1 and n+1

term therefore (X+1)^2=(1)/(2)X *2(X+3)

therefore x^2+2x+1=X(X+3) therefore 2x+1=3 thus X=1

therefore terms in order are 1/2*1;1+1=2;2(X+3)=8.1/2 ;2;8 are value of terms and

general term is 1/2(4)^n-1

here ratio r =4 that is obviously seen.

Assuming simplest condition n=1;2;3 is the order of given terms.

Answer 2

Given,

$\[\frac{1}{2}x,x + 1,2\left( {x + 3} \right)\]$

Solution,

Consider a, b, c are in G.P. then,

$\[{b^2} = ac\]$

Calculate the value of x,

Know that given terms are in G.P.

$\[{\left( {x + 1} \right)^2} = \frac{1}{2}x \times 2\left( {x + 3} \right)\]$

$\[\begin{array}{l} \Rightarrow {x^2} + 2x + 1 = x\left( {x + 3} \right)\\ \Rightarrow {x^2} + 2x + 1 = {x^2} + 3x\\ \Rightarrow x = 1\end{array}\]$

Hence the value of x is 1.

Calculate the value of general terms,

$\[\begin{array}{l}\frac{1}{2}x = \frac{1}{2} \times 1 = \frac{1}{2},\\x + 1 = 1 + 1 = 2\\2\left( {x + 3} \right) = 2 \times \left( {1 + 3} \right) = 8\end{array}\]$

Hence the sequence is $\[\frac{1}{2},2,8, - - - - \]$