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Question: Let m and n \(\left( m< n \right)\)be the roots of the equation \({{x}^{2}}-16x+39=0\). If four term...

Let m and n (m<n)\left( m< n \right)be the roots of the equation x2−16x+39=0{{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

Explanation

Solution

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
roots of the equation ax2+bx+c=0 are −b±b2−4ac2a.......................(1.1) \begin{aligned} & \text{roots of the equation }a{{x}^{2}}+bx+c=0\text{ are} \\\ & \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.......................(1.1) \\\ \end{aligned}
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=−(−16)−(−16)2−4×1×392=16−1002=3m=\dfrac{-(-16)-\sqrt{{{(-16)}^{2}}-4\times 1\times 39}}{2}=\dfrac{16-\sqrt{100}}{2}=3 and
n=−(−16)+(−16)2−4×1×392=16+1002=13n=\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
an=a0+(n−1)d{{a}_{n}}={{a}_{0}}+(n-1)d
Where a0{{a}_{0}} is the first term and d is the common difference. Comparing with equation (1.2), we get
a0=3...........(1.3){{a}_{0}}=3...........(1.3) and
a6=a0+(6−1)d⇒13=3+5d (from equation 1.2 and 1.3) ⇒d=13−35=2......................(1.4) \begin{aligned} & {{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{aligned}
Therefore, from equations (1.3) and (1.4), the terms p, q, r and s will be given by
p=a0+d=3+2=5 q=a0+2d=3+2×2=7 r=a0+3d=3+3×2=9 q=a0+4d=3+4×2=11..................(1.5) \begin{aligned} & 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{aligned}
Hence, from equation (1.5), we obtain
p+q+r+s=5+7+9+11=32p+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.