Question

If $\dfrac{\left( 5+9+13+...n\text{ terms} \right)}{\left( 7+9+11+...12\text{ terms} \right)}=\dfrac{5}{12}$, then n equals?
(a) 5
(b) 6
(c) 9
(d) 12

Answer 1

Hint: Consider the terms of the numerator and the denominator to be in AP. Use the formula for the sum of n terms of an AP given as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where a is the first term, d is the common difference and n is the number of terms. Use the values a = 5, d = 4 and n = n for the numerator and a = 7, d = 2 and n = 12 for the denominator. For a quadratic equation in n and use the middle term split method to get the answer.

Complete step-by-step solution:
Here we have been provided with the expression $\dfrac{\left( 5+9+13+...n\text{ terms} \right)}{\left( 7+9+11+...12\text{ terms} \right)}=\dfrac{5}{12}$ and we are asked to find the value of n.
Clearly we can see that both the numerator and the denominator are in AP as the difference between their consecutive terms is equal. In the numerator the difference between the consecutive terms is equal to 4 and in the denominator the difference is equal to 2. We know that the sum of n terms of an AP is given as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where a is the first term, d is the common difference and n is the number of terms.
(i) Considering the numerator $\left( 5+9+13+...n\text{ terms} \right)$ we have a = 5, d = 4 and n = n, so we get,
$\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2\times 5+\left( n-1 \right)\times 4 \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 10+4n-4 \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\times \left[ 6+4n \right] \\
\end{align}$
(ii) Considering the denominator $\left( 7+9+11+...12\text{ terms} \right)$ we have a = 7, d = 2 and n = 12, so we get,
$\begin{align}
  & \Rightarrow {{S}_{12}}=\dfrac{12}{2}\left[ 2\times 7+\left( 12-1 \right)\times 2 \right] \\
 & \Rightarrow {{S}_{12}}=\dfrac{12}{2}\left[ 14+11\times 2 \right] \\
 & \Rightarrow {{S}_{12}}=\dfrac{12}{2}\times \left[ 36 \right] \\
\end{align}$
Substituting the values of sum of the progressions in the given expression we get,
\[\Rightarrow \dfrac{\dfrac{n}{2}\times \left[ 6+4n \right]}{\dfrac{12}{2}\times \left[ 36 \right]}=\dfrac{5}{12}\]
Cancelling the common factors and simplifying by cross multiplication we get,
\[\begin{align}
  & \Rightarrow \dfrac{n\times \left[ 6+4n \right]}{36}=\dfrac{5}{1} \\
 & \Rightarrow 4{{n}^{2}}+6n=180 \\
 & \Rightarrow 2{{n}^{2}}+3n-90=0 \\
\end{align}\]
Using the middle term split method to factorize the quadratic equation in n we get,
\[\begin{align}
  & \Rightarrow 2{{n}^{2}}+15n-12n-90=0 \\
 & \Rightarrow \left( n-6 \right)\left( 2n+15 \right)=0 \\
\end{align}\]
Substituting each term equal to 0 we get,
\[\Rightarrow \left( n-6 \right)=0\] or \[\left( 2n+15 \right)=0\]
\[\Rightarrow n=6\] or \[n=\dfrac{-15}{2}\]
Since the number of terms cannot be negative or a fraction so $n=\dfrac{-15}{2}$ must be rejected. Therefore, we have the value n = 6.
Hence, option (b) is the correct answer.

Note: Do not forget to reject the negative or fractional value of n as it is invalid. The number of terms will always be a positive integer. You can also use the discriminant formula to solve the quadratic equation in n if you face difficulty in factoring the quadratic polynomial. Remember the formula for the sum of n terms of an AP.