Question

Find three consecutives even numbers such that the sum of first and last term exceeds the second one by 10.

Answer 1

Hint: We first try to define three consecutives even numbers which has to be in their general form. We then try to apply the given condition and make a linear equation out of it. We solve the equation and get the value of the middle term to find the other terms also.

Complete step by step answer:
We first need to define three consecutives even numbers. We first take the middle even number as $2k,k\in \mathbb{N}$.
The previous and the next even numbers will be $2k-2$ and $2k+2$.
We have been given that the sum of the first and last term exceeds the second one by 10 which means the sum of $2k-2$ and $2k+2$ exceeds the 2k by 10.
Sum of $2k-2$ and $2k+2$ will be \[\left( 2k-2 \right)+\left( 2k+2 \right)=4k\]. This exceeds 2k by 10.
This gives us $4k=2k+10$. Solving the equation, we get
$\begin{align}
  & 4k=2k+10 \\
 & \Rightarrow 2k=10 \\
\end{align}$
So, the middle term is 10. The other two terms are $2k-2=10-2=8$ and $2k+2=10+2=12$.

Therefore, the terms are 8, 10, 12..

Note: Instead of taking the first term we tried to take the middle term as the main variable because the sum of first and last term eliminates the constants to get to the variable’s value easily. In the case of A.P. this type of assumption helps a lot to solve the problem easily.