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Find the sum of all integers between $84$ and $719$ , which are multiples of $5$ .

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 Hint: We have to find the sum between $84$ and $719$ , which are multiples of $5$ . We can see it is A.P, so the series will be $85,90,.....,715$. Now we have got the terms we want to find How much number of terms are there, so for that use ${{a}_{n}}={{a}_{1}}+(n-1)d$ , you will get the number of terms and after that, use ${{S}_{n}}=\dfrac{n}{2}({{a}_{1}}+{{a}_{n}})$ to find the sum. Try it, you will get the answer.

Complete step-by-step Solution:

Like the natural numbers, Z is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers, and, importantly, 0, Z (unlike the natural numbers) is also closed under subtraction. The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring Z.
Z is not closed under division, since the quotient of two integers (e.g., 1 divided by 2), need not be an integer. Although the natural numbers are closed under exponentiation, the integers are not (since the result can be a fraction when the exponent is negative).
The integers form the smallest group and the smallest ring containing the natural numbers. In algebraic number theory, the integers are sometimes qualified as rational integers to distinguish them from the more general algebraic integers. In fact, the (rational) integers are the algebraic integers that are also rational numbers.
Integers are whole numbers and their negative opposites.
A number with no fractional part (no decimals).
In Maths, integers are the numbers which can be positive, negative or zero. These numbers are used to perform various arithmetic calculations, like addition, subtraction, multiplication and division.
All integers between $84$ and $719$ , which are multiple of $5$ are $85,90,.....,715$.
So this is in A.P.
First term$={{a}_{1}}=85$ and last term$={{a}_{n}}=715$.
Common difference$=d=90-85=5$.
So ${{n}^{th}}$ term of A.P is given by,
${{a}_{n}}={{a}_{1}}+(n-1)d$
$\begin{align}
  & 715=85+(n-1)5 \\
 & 715-85=(5n-5) \\
 & 630=5n-5 \\
 & 630+5=5n \\
 & \dfrac{635}{5}=n \\
 & 127=n \\
\end{align}$
We get $127$ number of terms.
Now sum of this $127$ terms is,
${{S}_{n}}=\dfrac{n}{2}({{a}_{1}}+{{a}_{n}})$
$\begin{align}
  & {{S}_{127}}=\dfrac{127}{2}(85+715) \\
 & {{S}_{127}}=\dfrac{127}{2}(800) \\
 & {{S}_{127}}=50800 \\
\end{align}$
Hence, the sum of all integers from $84$ to $719$ is $50800$ .

Note: Read the question carefully. Do not make silly mistakes. Don’t get confused while solving the problem. Your concept regarding A.P should be clear. Do not jumble yourself while simplifying.