Question

If a, b, c be the $ {{4}^{th}},{{7}^{th}} $ and $ {{10}^{th}} $ term of an AP respectively, then the sum of the roots of the equation $ a{{x}^{2}}-2bx+c=0 $ is
a. $ -\dfrac{b}{a} $
b. $ -\dfrac{2b}{a} $
c. $ \dfrac{c+a}{a} $
d. Cannot be determined unless some more information is given about the AP.

Answer 1

Hint:In in order to find the solution of this question, we will first find the values of a, b and c by the formula of finding $ {{n}^{th}} $ term of AP, that is, $ {{a}_{n}}=a+\left( n-1 \right)d $ then we will use those values in the given quadratic equation and then apply middle term split method to get the roots of the equation and then we will find their sum.

Complete step-by-step answer:
In this question, we have been asked to find the sum of the roots of the quadratic equation, $ a{{x}^{2}}-2bx+c=0 $ , where a, b and c are the $ {{4}^{th}},{{7}^{th}} $ and $ {{10}^{th}} $ term of an AP respectively.
Now, we know that the $ {{n}^{th}} $ term of AP, whose first term is $ {{a}_{1}} $ and common difference is d, is calculated by the formula, $ {{a}_{n}}=a+\left( n-1 \right)d $ . So, for the given AP, let us consider $ {{a}_{1}} $ as the first term and d as the common difference, so we can say that, $ {{a}_{n}}={{a}_{1}}+\left( n-1 \right)d $ . Now, we know that the $ {{4}^{th}} $ term is given as a. So, we can say, $ {{a}_{4}}={{a}_{1}}+\left( 4-1 \right)d\Rightarrow a={{a}_{1}}+3d $ . Also, we have been given that the $ {{7}^{th}} $ term is given as b. So, we can say, $ {{a}_{7}}={{a}_{1}}+\left( 7-1 \right)d\Rightarrow b={{a}_{1}}+6d $ . And we have been also given that the $ {{10}^{th}} $ term is given as c. So, we can say that, $ {{a}_{10}}={{a}_{1}}+\left( 10-1 \right)d\Rightarrow c={{a}_{1}}+9d $ .
So, by using these values of a, b and c, we can say that the quadratic equation will become,
\[\begin{align}
  & \left( {{a}_{1}}+3d \right){{x}^{2}}-2\left( {{a}_{1}}+6d \right)x+\left( {{a}_{1}}+9d \right)=0 \\
 & {{a}_{1}}{{x}^{2}}+3d{{x}^{2}}-2{{a}_{1}}x-12dx+{{a}_{1}}+9d=0 \\
 & {{a}_{1}}\left( {{x}^{2}}-2x+1 \right)+3d\left( {{x}^{2}}-4x+3 \right)=0 \\
\end{align}\]
Now, we will apply the quadratic equation we obtained. So, we get,
\[\begin{align}
  & {{a}_{1}}\left( {{x}^{2}}-x-x+1 \right)+3d\left( {{x}^{2}}-3x-x+3 \right)=0 \\
 & {{a}_{1}}\left( x\left( x-1 \right)-1\left( x-1 \right) \right)+3d\left( x\left( x-3 \right)-1\left( x-3 \right) \right)=0 \\
 & {{a}_{1}}\left( x-1 \right)\left( x-1 \right)+3d\left( x-3 \right)\left( x-1 \right)=0 \\
 & \left( x-1 \right)\left[ {{a}_{1}}\left( x-1 \right)+3d\left( x-3 \right) \right]=0 \\
 & \left( x-1 \right)\left[ {{a}_{1}}x-{{a}_{1}}+3dx-9d \right]=0 \\
\end{align}\]
To satisfy this equality, either of the equations should be zero. So, we can say,
x - 1 = 0 or \[{{a}_{1}}x-{{a}_{1}}+3dx-9d=0\]
x = 1 or \[\left( {{a}_{1}}+3d \right)x-\left( {{a}_{1}}+9d \right)=0\]
x = 1 or \[x=\dfrac{{{a}_{1}}+9d}{{{a}_{1}}+3d}\]
Now, we have been asked to calculate the sum of the roots of the equation, $ a{{x}^{2}}-2bx+c=0 $ . So, let us consider $ x=1=\alpha $ and \[x=\dfrac{{{a}_{1}}+9d}{{{a}_{1}}+3d}=\beta \]. Therefore, we get,
 $ \alpha +\beta =1+\dfrac{{{a}_{1}}+9d}{{{a}_{1}}+3d} $
Now, we know that $ {{a}_{1}}+9d=c $ and $ {{a}_{1}}+3d=a $ . So, we can say,
 $ \begin{align}
  & \alpha +\beta =1+\dfrac{c}{a} \\
 & \alpha +\beta =\dfrac{a+c}{a} \\
 & \alpha +\beta =\dfrac{c+a}{a} \\
\end{align} $
Hence, we can say that for the quadratic equation, $ a{{x}^{2}}-2bx+c=0 $ where a, b, c are the $ {{4}^{th}},{{7}^{th}} $ and $ {{10}^{th}} $ term of an AP respectively, then the sum of the roots of the equation is $ \dfrac{c+a}{a} $ .
Hence, option (c) is the correct answer.

Note: While solving this question, the possible mistake we can make is by choosing option (b) as the correct answer which is wrong because we know that the sum of roots for an equation $ a{{x}^{2}}+bx+c=0 $ is given by $ -\dfrac{b}{a} $ . So, for $ a{{x}^{2}}-2bx+c=0 $ , we get the sum of roots as $ \dfrac{2b}{a} $ not $ -\dfrac{2b}{a} $ . Therefore, we have to be careful while selecting the correct answer.