Question

The square of which of the following numbers is the difference between the ${49^{th}}$ even number after $1575$ and ${70^{th}}$even number before $1028$.

Answer 1

In this question, we are asked to find out the square of the difference between the ${49^{th}}$ even number after $1575$ and ${70^{th}}$ even number before $1028$.
To solve this question, we are going to use the concept of arithmetic progression.
In arithmetic progression, we will be using the nth term and last term formula to obtain the required even number.
Formula used:
Here we will use the given formula
${a_n} = a + \left( {n - 1} \right)d$

Complete step by step solution:
We know that ${49^{th}}$ even number after $1575$ will be the same as ${49^{th}}$ even number after $1574$
Using the concept of arithmetic progression, we get
The 49th even number after is: $1672$
First even number\left( a \right)$$$$ = $$$$1576,d$$$$ = $$$$2
Hence the ${n^{th}}$in an A.P is defined as:
${a_n} = a + \left( {n - 1} \right)d$
Here, n$$$$ = $$$$49 and a$$$$ = $$$$1576
Last term$\left( {{a_{49}}} \right) = a + \left( {49 - 1} \right)d$
${a_{49}} = a + 48d$
${a_{49}} = 1576 + 48 \times 2$j
${a_{49}} = 1672$
${70^{th}}$even number is also done by concept of arithmetic progression
${a_n} = a + \left( {n - 1} \right)d$
$= 888$
Hence the difference between $1672$and $888$ is
$1672 - 888 = 784$
$\sqrt {784} = 28$
Hence, $28$ is the correct answer.
Therefore, the square of the difference between the ${49^{th}}$ even number after $1575$ and ${70^{th}}$ even number before $1028$ is $28$.

Note: Students must avoid calculation mistakes to get the correct answers. Calculation mistakes often lead them to an incorrect answer.
Students must remember the formula of arithmetic progression.
Alternate method:
Let the number be $x$
${49^{th}}$ even number after $1575$ will be same as ${49^{th}}$ even number after $1574$
$= 1574 + 2 \times 49$
$= 1574 + 98$
$= 1672$
${70^{th}}$even number before $1028$ will be $888$
Given,
${x^2} = 1672 - 888$
${x^2} = 784$
$x = \sqrt {784}$
$= 28$
Therefore, we get the same answer from this method also. Students can apply any of the methods to get the required answer.
Students should also remember the last term formula which is given by ${a_l} = l - \left( {n - 1} \right)d$