Question

If a, ${{a}^{2}}+2$ and ${{a}^{3}}+10$ are in GP, then find the value(s) of a.

Answer 1

Here, we may apply the concept that a GP is generated by multiplying each of its terms by a constant every time to get its successive terms and this constant is called the common ratio of the GP.

Complete step-by-step answer:

Since, the given terms of the GP are:

$a,\,{{a}^{2}}+2\,\,and\,{{a}^{3}}+10$.

We know that when the first term of the GP that is a is multiplied by the common ratio, we get the second term of the GP.

Let the common ratio of this GP be = r.

So, the common ratio will be:

$r=\dfrac{{{a}^{2}}+2}{a}...........(1)$

Similarly when we multiply the second term of the GP with the common ratio, then we will get the third term.

So, again for the common ratio we have:

$r=\dfrac{{{a}^{3}}+10}{{{a}^{2}}+2}...............(2)$

Since, both the equations (1) and (2) give a common ratio of the GP. So, we can equate both the equations and get:

$\dfrac{{{a}^{2}}+2}{a}=\dfrac{{{a}^{3}}+10}{{{a}^{2}}+2}$

On cross multiplication and further simplification we will get:

$\left( {{a}^{2}}+2 \right)\times \left( {{a}^{2}}+2 \right)=\left( {{a}^{3}}+10 \right)\times a$

${{a}^{4}}+2{{a}^{2}}+2{{a}^{2}}+4={{a}^{4}}+10a$

$4{{a}^{2}}-10a+4=0$

So, we have a quadratic equation in ‘a’. It means that by solving this equation we may have two values of a. So on dividing the equation so obtained by 2 we get:

$2{{a}^{2}}-5a+2=0...........(3)$

Now, we may solve this equation using factorization method:

$2{{a}^{2}}-4a-a+2=0$

$2a\left( a-2 \right)-1\left( a-2 \right)=0$

$\left( 2a-1 \right)\left( a-2 \right)=0$

So, we have:

$2a-1=0\,\,\,\,\,\,or\,\,\,a-2=0$

Hence, the two values of a are:

$a=\dfrac{1}{2}\,\,\,or\,\,\,a=2$

Note: Students should note here that the common ratio of a GP is always given as a term divided by its preceding term. So, there is a chance of mistake if we are dividing a term to its succeeding term.