Question

Let m and n $\left( m< n \right)$be the roots of the equation ${{x}^{2}}-16x+39=0$. If four terms p, q, r and s are inserted between m and n and form an AP, then what is the value of p+q+r+s?
a)29
b)30
c)32
d)35

Answer 1

Hint: In this case, we are given to find the sum of p, q, r and s which are inserted between m and n to form an AP. Therefore, first we should solve the quadratic equation to find m and n and then find the form of the AP created by m, p, q, r, s and n.

Complete step-by-step answer:
In this case, to find the roots of the given quadratic equation, we use the formula for finding the roots of the general quadratic equation as
$\begin{align}
  & \text{roots of the equation }a{{x}^{2}}+bx+c=0\text{ are} \\
 & \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.......................(1.1) \\
\end{align}$
i.e. taking + and – sign gives us the two roots of the equation.
Compared to equation (1.1), in this case, a=1, b= -16 and c=39. Therefore, the roots are
\[m=\dfrac{-(-16)-\sqrt{{{(-16)}^{2}}-4\times 1\times 39}}{2}=\dfrac{16-\sqrt{100}}{2}=3\] and
\[n=\dfrac{-(-16)+\sqrt{{{(-16)}^{2}}-4\times 1\times 39}}{2}=\dfrac{16+\sqrt{100}}{2}=13\]
Now, as given in the question the sequence m, p, q, r, s and n forms an arithmetic progression. Therefore, in this AP the first term is m=3 and the sixth term is n=13……………………. (1.2)
As, in an AP the nth term is given by
${{a}_{n}}={{a}_{0}}+(n-1)d$
Where ${{a}_{0}}$ is the first term and d is the common difference. Comparing with equation (1.2), we get
${{a}_{0}}=3...........(1.3)$ and
$\begin{align}
  & {{a}_{6}}={{a}_{0}}+(6-1)d\Rightarrow 13=3+5d\text{ (from equation 1}\text{.2 and 1}\text{.3)} \\
 & \Rightarrow \text{d=}\dfrac{13-3}{5}=2......................(1.4) \\
\end{align}$
Therefore, from equations (1.3) and (1.4), the terms p, q, r and s will be given by
$\begin{align}
  & p={{a}_{0}}+d=3+2=5 \\
 & q={{a}_{0}}+2d=3+2\times 2=7 \\
 & r={{a}_{0}}+3d=3+3\times 2=9 \\
 & q={{a}_{0}}+4d=3+4\times 2=11..................(1.5) \\
\end{align}$
Hence, from equation (1.5), we obtain
$p+q+r+s=5+7+9+11=32$
Which matches option (c). Hence, option c is the correct answer to this question.

Note: After obtaining the values of p, q, r and s, we could also have found their sum by considering them to be an AP with first term p and common difference 2. However, the obtained answer would be the same as obtained in the solution above.