Question

Find the smallest number of three numbers constituting a GP if it is known that the sum of the numbers is equal to 26 and that when 1, 6 and 3 are added to them respectively, then the new numbers that are obtained form an AP.

Answer 1

Hint: Assume a variable a which will represent the first term of the GP and assume a variable r which will represent the common ratio of this GP. The sum for n terms of this GP is given by the formula $S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$. Also, if three numbers a, b and c are in AP, then we can say that 2b = a + c. Using this, we can solve this question.


Complete step-by-step answer:
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In sequences and series, if we have a GP with first term as a and the common ratio as r, then the sum of the n terms of this GP is given by the formula,
$S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ . . . . . . . . . . . . . (1)

Also, in sequences and series, if we are given that the three numbers a, b and c are in AP, the we can say that,
2b = a + c . . . . . . . . . . . . . (2)

In the question, we are given a GP having three terms and the sum of these three terms is equal to 26. Let us assume that the first term is a and the common ratio is r. Substituting S = 26 and n = 3 in formula (1), we get,
$26=\dfrac{a\left( {{r}^{3}}-1 \right)}{r-1}$ . . . . . . . . . . . . (3)
Also, it is given that when we add 1, 6 and 3 to these terms respectively, then the new numbers obtained form an AP. So, we can say,
a+1, ar + 6 and $a{{r}^{2}}$ + 3 are in AP. Using formula (2), we get,
$\begin{align}
 & 2\left( ar+6 \right)=\left( a+1 \right)+\left( a{{r}^{2}}+3 \right) \\
 & \Rightarrow 2ar+12=a+a{{r}^{2}}+4 \\
 & \Rightarrow 2ar+8=a+a{{r}^{2}} \\
 & \Rightarrow a\left( {{r}^{2}}-2r+1 \right)=8 \\
 & \Rightarrow a{{\left( r-1 \right)}^{2}}=8 \\
 & \Rightarrow a=\dfrac{8}{{{\left( r-1 \right)}^{2}}} \\
\end{align}$

Substituting this value of a in equation (3), we get,
\[\begin{align}
 & 26=\dfrac{8\left( {{r}^{3}}-1 \right)}{{{\left( r-1 \right)}^{3}}} \\
 & \Rightarrow 26{{\left( r-1 \right)}^{3}}=8\left( {{r}^{3}}-1 \right) \\
 & \Rightarrow 26\left( r-1 \right){{\left( r-1 \right)}^{2}}=8\left( r-1 \right)\left( {{r}^{2}}+r+1 \right).....................\left( 4 \right) \\
 & \Rightarrow 26\left( {{r}^{2}}-2r+1 \right)=8\left( {{r}^{2}}+r+1 \right) \\
 & \Rightarrow 26{{r}^{2}}-52r+26=8{{r}^{2}}+8r+8 \\
 & \Rightarrow 18{{r}^{2}}-60r+18=0 \\
 & \Rightarrow 6\left( 3{{r}^{2}}-10r+3 \right)=0 \\
 & \Rightarrow 3{{r}^{2}}-10r+3=0 \\
\end{align}\]

For a quadratic equation $a{{x}^{2}}+bx+c=0$, the roots of this quadratic equation are given by the formula x = $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Using this formula, the roots of the above quadratic equation is,
\[\begin{align}
 & r=\dfrac{-\left( -10 \right)\pm \sqrt{{{\left( -10 \right)}^{2}}-4.3.3}}{2.3} \\
 & \Rightarrow r=\dfrac{10\pm \sqrt{100-36}}{6} \\
 & \Rightarrow r=\dfrac{10\pm \sqrt{64}}{6} \\
 & \Rightarrow r=\dfrac{10\pm 8}{6} \\
 & \Rightarrow r=3,r=\dfrac{1}{3} \\
\end{align}\]

Since $a=\dfrac{8}{{{\left( r-1 \right)}^{2}}}$, we get $a=2,a=18$
Since the GP is $a,ar,a{{r}^{2}}$, we can get two possible GP that are,
$2,6,18$ or $18,6,2$.
In both the GP, the smallest term is 2.
Hence, the answer is 2.

Note: In the question, it is given that we have to find the smallest term of the GP. This means, we can assume that all the three terms of the GP must be different i.e. the common ratio $r\ne 1$. That is the reason why we cancelled the term (r-1) on both the sides in equation (4) since $r\ne 1$.