Question

If the arithmetic mean and geometric mean of the $pth$ and $qth$ terms of the sequence $-16,8,-4,2,\dots .$ satisfy the equation $4x^2-9x+5=0$, then $p+q$ is equal to

Answer 1

Step 1: Find the roots of equation.

Given equation is : $4x^2-9x+5=0$

By middle term splitting method we have:

$\begin{array}{rcl}4x^2-9x+5& =& 0\\ 4x^2-4x-5x+5& =& 0\\ 4x\left(x-1\right)-5\left(x-1\right)& =& 0\\ \left(x-1\right)\left(4x-5\right)& =& 0\end{array}$

$\Rightarrow x=1,\frac{5}{4}$

Since the $pth$ and $qth$ terms of the sequence $-16,8,-4,2,\dots .$satisfies the equation it means that $pth$ and $qth$ term are the roots of equation $4x^2-9x+5=0$

so, $A.Mofpthandqthterms=\frac{5}{4}\mathrm{and}\mathrm{G}.\mathrm{M}\mathrm{of}\mathrm{pth}\mathrm{and}\mathrm{qth}\mathrm{terms}=1$ [since,$A.M\ge G.M$]

Step 2: Find the terms of the sequence

As the sequence is in Geometric progression so the terms are:

$\begin{array}{rcl}{t}_p& =& -16{\left(-\frac{1}{2}\right)}^{p-1}\\ & =& (-1{)}^p{2}^{5-p}\end{array}$

$\begin{array}{rcl}{t}_q& =& -16{\left(\frac{-1}{2}\right)}^{q-1}\\ & =& (-1{)}^q{2}^{5-q}\end{array}$
Since, the geometric mean is $1$.

So,

$\begin{array}{rcl}\sqrt{(-1{)}^p{2}^{5-p}\times (-1{)}^q{2}^{5-q}}& =& 1\end{array}$
$\begin{array}{rcl}\Rightarrow \sqrt{(-1{)}^{p+q}{2}^{10-p-q}}& =& 1\end{array}$

$\begin{array}{rcl}\Rightarrow (-1{)}^{p+q}{2}^{10-p-q}& =& 2^0\end{array}$

$\begin{array}{rcl}& \Rightarrow & 10-p-q=0\\ & \Rightarrow & p+q=10\\ & & \end{array}$

Hence, the value of $p+q=10$.