
What is the sum of 2-digit multiples of $4$?
A) $998$
B) $1188$
C) $1298$
D) $1388$
Answer
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Hint: In this question, we have been asked the sum of all the 2-digit multiples of 4. This question can be done in many ways but we will use arithmetic progression to find the answer to this question. At first, we will find the smallest and the largest 2-digit number divisible by 2. Now, we will use the formula of sum of A.P involving first and last term. But for this, we will also require the number of terms. We will use the formula of the general term of A.P to find the number of terms.
Formula used: 1) ${a_n} = a + \left( {n - 1} \right)d$
2) ${S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$
Complete step-by-step solution:
We will start by finding the 2-digit numbers divisible by 4.
At first, we will find the smallest 2-digit number divisible by 4. If we look at the table of 4, we will notice that 12 is the smallest 2-digit number divisible by 4.
Next, we will find the largest 2-digit number divisible by 4. We know that 100 is divisible by 4. If we subtract 4 from 100, we will get the largest 2-digit number divisible by 4. $ \Rightarrow 100 - 4 = 96$
Hence, 96 is our required number.
Now, all the 2-digit numbers divisible by 4 would lie between 12 and 96. But how many numbers are there? To find out this, we will use the general formula of A.P $ \Rightarrow {a_n} = a + \left( {n - 1} \right)d$.
Putting $a = 12,{a_n} = 96$ and our d will be 4 as all the numbers are divisible by 4.
$ \Rightarrow 96 = 12 + \left( {n - 1} \right)4$
Solving to find $'n'$,
$ \Rightarrow 96 - 12 = \left( {n - 1} \right)4$
$ \Rightarrow \dfrac{{84}}{4} + 1 = n$
$ \Rightarrow n = 22$
Now, we have all the required values. We will use the formula of sum of A.P to find the sum.
$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$
$ \Rightarrow {S_n} = \dfrac{{22}}{2}\left( {12 + 96} \right)$
Solving to find the sum,
$ \Rightarrow {S_n} = 11 \times 108$
$ \Rightarrow {S_n} = 1188$
Hence, the sum of all the 2-digit multiples of 4 is option (B) 1188.
Note: We already know that in mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance the sequence 5, 7, 9, 11, 13, 15, …… is an arithmetic progression with a common difference of 2. A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of finite arithmetic progression is called an arithmetic series.
1) In the formula, ${a_n} = a + \left( {n - 1} \right)d$, ${a_n}$= $n^{th}$ term, $a = $ first term, $n = $number of terms, $d = $ common difference.
2) In the formula, ${S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$, $n = $ number of terms, $a = $ first term, ${a_n} = $ last term.
Formula used: 1) ${a_n} = a + \left( {n - 1} \right)d$
2) ${S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$
Complete step-by-step solution:
We will start by finding the 2-digit numbers divisible by 4.
At first, we will find the smallest 2-digit number divisible by 4. If we look at the table of 4, we will notice that 12 is the smallest 2-digit number divisible by 4.
Next, we will find the largest 2-digit number divisible by 4. We know that 100 is divisible by 4. If we subtract 4 from 100, we will get the largest 2-digit number divisible by 4. $ \Rightarrow 100 - 4 = 96$
Hence, 96 is our required number.
Now, all the 2-digit numbers divisible by 4 would lie between 12 and 96. But how many numbers are there? To find out this, we will use the general formula of A.P $ \Rightarrow {a_n} = a + \left( {n - 1} \right)d$.
Putting $a = 12,{a_n} = 96$ and our d will be 4 as all the numbers are divisible by 4.
$ \Rightarrow 96 = 12 + \left( {n - 1} \right)4$
Solving to find $'n'$,
$ \Rightarrow 96 - 12 = \left( {n - 1} \right)4$
$ \Rightarrow \dfrac{{84}}{4} + 1 = n$
$ \Rightarrow n = 22$
Now, we have all the required values. We will use the formula of sum of A.P to find the sum.
$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$
$ \Rightarrow {S_n} = \dfrac{{22}}{2}\left( {12 + 96} \right)$
Solving to find the sum,
$ \Rightarrow {S_n} = 11 \times 108$
$ \Rightarrow {S_n} = 1188$
Hence, the sum of all the 2-digit multiples of 4 is option (B) 1188.
Note: We already know that in mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance the sequence 5, 7, 9, 11, 13, 15, …… is an arithmetic progression with a common difference of 2. A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of finite arithmetic progression is called an arithmetic series.
1) In the formula, ${a_n} = a + \left( {n - 1} \right)d$, ${a_n}$= $n^{th}$ term, $a = $ first term, $n = $number of terms, $d = $ common difference.
2) In the formula, ${S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$, $n = $ number of terms, $a = $ first term, ${a_n} = $ last term.
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