Question

A group of friends have some money which was in an increasing GP. The total money with the first and the last friend was Rs. 66 and the product of the money that the second friend has and that the last but one friend has was Rs. 128. If the total money with all of them together was Rs. 126, How many friends were there?
A) 6
B) 5
C) 3
D) Cannot be determined

Answer 1

Hint:
Here we will consider an increasing GP $a,ar,a{r}^2,.......a{r}^{n-1}$. Then from the given question we will try to form equations. We will have 2 equations. Then we will put the value of equation 2 in equation 1. Then we will get the value of ‘a’. Let’s see how we can solve it.

Complete step by step solution:
Let’s consider the GP $a_1,a_2,a_{3,}......a_n$
We know that $a_n=a{r}^{n-1}$
Putting the value n = 1 in the above formula;
$a_1=a{r}^{1-1}$
$\Rightarrow a_1=a$
Similarly, we can say that the increasing GP will be $a,ar,a{r}^2,.......a{r}^{n-1}$.
Sum of 1st term and the last term is a + $a{r}^{n-1}$= 66 $\Rightarrow a(1+r^{n-1})=66$……. (1)

Product of second and last second term = $(ar)(a{r}^{n-2})=128$
= $a^2{r}^{n-2+1}=128$
= $a^2{r}^{n-1}=128$
= $r^{n-1}={\displaystyle \frac{128}{a^2}}$……. (2)

Putting the value $r^{n-1}={\displaystyle \frac{128}{a^2}}$in equation (1), we get;
$a(1+r^{n-1})=66$
$\Rightarrow a\left(1+{\displaystyle \frac{128}{a^2}}\right)=66$
Solving the equation for a; we get
$$\begin{gathered} \Rightarrow a^2+128=66a \\ \Rightarrow a^2-66a+128=0 \end{gathered}$$
Factoring the above quadratic equation, we get;
$$\begin{gathered} \Rightarrow a^2-64a-2a+128=0 \\ \Rightarrow a(a-64)-2(a-64)=0 \end{gathered}$$
$\Rightarrow$(a – 64) (a – 2) = 0
$\Rightarrow$a = 64 and a = 2

It is not possible to have 2 first terms. So, we will put the values of a in
$r^{n-1}={\displaystyle \frac{128}{a^2}}$
Putting the value, of a = 64
$r^{n-1}={\displaystyle \frac{128}{{(64)}^2}}={\displaystyle \frac{1}{32}}$
$r^n={\displaystyle \frac{1}{32}}r$
Now, Putting the value, a = 2
$r^{n-1}={\displaystyle \frac{128}{a^2}}$
$r^{n-1}={\displaystyle \frac{128}{{(2)}^2}}={\displaystyle \frac{128}{2\times 2}}=32$
$\Rightarrow r^{n-1}=32$………… (3)
$$\begin{gathered} \Rightarrow r^n.r^{-1}=32 \\ \Rightarrow r^n=32r \end{gathered}$$
Clearly it is mentioned in the question that we have increasing GP. Thus, in increasing GP we cannot consider the value of $r^n$ which is less than 1 as it will not be an increasing GP anymore. So, we cannot take $r^n$ = ${\displaystyle \frac{1}{32}}r$. That is why we would consider a = 2 and $r^n$= 32r
Sum of total money they have $S_n$ = 126
In GP, $S_n={\displaystyle \frac{a(r^n-1)}{r-1}}$
$S_n$= ${\displaystyle \frac{2(32r-1)}{r-1}}=126$
Solving for r;
$\Rightarrow$32r – 1 = 63(r – 1)
$\Rightarrow$32r – 1 = 63r – 63
$\Rightarrow$ 31r – 62 = 0
$\Rightarrow$r = ${\displaystyle \frac{62}{31}}=2$
Now we know that from above that $r^{n-1}=32$
Putting the value of r = 2;
$r^{n-1}=32$
$\Rightarrow$$2^{n-1}=2^5$
Base is same, so we can equate their powers;
$$\begin{gathered} n–1=5 \\ \Rightarrow n=6 \end{gathered}$$
Thus, the total number of friends is 6.

Hence, Option A is the correct option.

Note:
In the above question, first you need to try to find out the value of a &$r^n$. You have to keep in mind that you have to find a number of friends. It means you need to find the value of ‘n’. It is given that$S_n={\displaystyle \frac{a(r^n-1)}{r-1}}$= 126. Now, you have to put the value of a & $r^n$in this equation. After solving this you can find the value of ‘r’. Now, you can put the value of ‘r’ in equation 3. You will finally get the value of ‘n’.