Question

If \[x\] is added to each of number \[3,9,21\] so that the resulting numbers may be in G.P, then the value of \[x\] will be
1)\[3\]
2)\[\dfrac{1}{2}\]
3)\[2\]
4)\[\dfrac{1}{3}\]

Answer 1

Hint: If geometric progression is known to you it will be very easy for you to solve this question. Firstly we will add a variable \[x\]to each number then we will find the geometric mean of the new series formed which will give us the value of\[x\].In this way we can obtain the required result of the given question.

Complete step-by-step solution:
There are basically three types of series: -
Arithmetic Series
Geometric progression
Harmonic Series
Harmonic series is just reciprocal to Arithmetic Series.

In a sequence of terms , each succeeding term is generated by multiplying each preceding term with a constant value then the sequence is called a geometric progression, where the constant value is called the common ratio.
The general form of Geometric progression is: \[a,ar,a{{r}^{2}},a{{r}^{3}},.....,a{{r}^{n-1}}\]
Where,
\[a\]= First term of the series
\[r\]= common ratio of the series
\[a{{r}^{n-1}}\]= ${{n}^{th}}$of the series
Now,
First term= \[a\]
Second term= \[ar\]
 Third term= \[a{{r}^{2}}\]
Thus, common ratio = (any term) divided by (preceding term)
So,
Common ratio\[=\dfrac{a{{r}^{2}}}{ar}\]
= \[r\]
Neither the common ratio nor any term of the G.P can be zero.
Now, according to question, if \[x\]is added to each number then new Geometric progression series becomes
\[3+x,9+x,21+x\]
If three terms are in G.P, then the middle one is called the geometric mean of the other two terms.
For example, if \[a,b,c\]are in G.P then \[b\] is the geometric mean of \[a\] and \[c\]. This can be written as \[{{b}^{2}}=ac\] or \[b=\sqrt{ac}\].
Now,
Apply the same concept on given question we get,
\[{{(9+x)}^{2}}=(3+x)(21+x)\]
On solving this we will get,
\[81+18x+{{x}^{2}}=63+24x+{{x}^{2}}\]
\[\Rightarrow 18=6x\]
Which gives us,
\[\Rightarrow x=3\]
So the required and is 3.
When we add 3 to the numbers given in the question, it would become\[6,12,24\], which is in G.P.
Where the first term that is \[a=6\]
Second term that is \[ar\]\[=12\]
Third term that is \[a{{r}^{2}}\]\[=24\]
And the common ratio can be found by \[=\dfrac{a{{r}^{2}}}{ar}\]
\[\Rightarrow \dfrac{24}{12}\]
\[\therefore 2\]
So the resulting answer is 2
So Correct Option is 3.

Note: If every term of Geometric progression is multiplied or divided by a non-zero number, then the resulting series is also in Geometric progression. If all the terms of Geometric progression are raised to the same power, then the resulting series is also in G.P. Reciprocal of all the terms in G.P also form a G.P.