Question

If $a,b,c$ be the $4^{\text{th}},7^{\text{th}}$ and ${10}^{\text{th}}$ term of an AP respectively then the sum of the roots of the equation $a{x}^2-2bx+c=0$:

• A. $-\frac{b}{a}$
• B. $-\frac{2b}{a}$
• C. $\frac{c+a}{a}$
• D. Can not be determined unless some more information is given about the AP

Answer 1

The correct option is C

$\frac{c+a}{a}$

Step 1: Given information.

It is given that, $a,b,c$ are the $4^{\text{th}},7^{\text{th}}$ and ${10}^{\text{th}}$ term of an AP respectively.

We know that the formula for the $n^{\text{th}}$ term of AP, whose first term is $a$ and the common difference is $d$ is given by

$a_n=a+\left(n-1\right)d$

For the given AP, let $a_1$ be the first team and $d$ be a common difference, so we have

$a_n=a_1+\left(n-1\right)d$

As the $4^{\text{th}}$ term is given as $a$. So, we can write as follows:

$$\begin{gathered} a_4=a_1+\left(4-1\right)d \\ \Rightarrow a=a_1+3d...\left(1\right) \end{gathered}$$

Similarly, the $7^{\text{th}}$ term is given as $b$. So, we can write as:

$$\begin{gathered} a_7=a_1+\left(7-1\right)d \\ \Rightarrow b=a_1+6d...\left(2\right) \end{gathered}$$

Also, the ${10}^{\text{th}}$ term is given as $c$. Therefore, we have

$$\begin{gathered} a_{10}=a_1+\left(10-1\right)d \\ \Rightarrow c=a_1+9d...\left(3\right) \end{gathered}$$

Step 2: Simplify the above equations to get the desired result.

Subtract equation $\left(1\right)$ from equation $\left(2\right)$ as follows:

$$\begin{gathered} b-a=\left(a_1+6d\right)-\left(a_1+3d\right) \\ \Rightarrow b-a=3d \\ \Rightarrow d=\frac{b-a}{3} \end{gathered}$$

Substitute the value of $d$ in equation $\left(1\right)$ and simplify.

$$\begin{gathered} a=a_1+3\left(\frac{b-a}{3}\right) \\ \Rightarrow a=a_1+b-a \\ \Rightarrow a_1=2a-b \end{gathered}$$

Substitute the value of $d$ and $a_1$ in equation $\left(3\right)$ and simplify as follows:

$$\begin{gathered} c=\left(2a-b\right)+9\left(\frac{b-a}{3}\right) \\ \Rightarrow c=2a-b+3b-3a \\ \Rightarrow c=2b-a \\ \Rightarrow 2b=c+a...\left(4\right) \end{gathered}$$

Step 3: Finding the sum of roots

Now, the sum of the roots of the quadratic equation $a{x}^2+bx+c=0$ is given by, $-\frac{b}{a}$.

So, for the given equation $a{x}^2-2bx+c=0$, the sum of the roots is $\frac{2b}{a}$.

From equation $\left(4\right)$, we get

$\frac{2b}{a}=\frac{c+a}{a}$

Hence, the option $\left(c\right)$ is correct.