Question

The sum of the numbers lying between $10$ and $200$ that are divisible by $7$ will be
a) $2800$
b) $2835$
c) $2870$
d) $2849$

Answer 1

We have to find the sum of the numbers that are divisible by $7$ and lie between $10$ and $200$. To solve this, we will use the concept of arithmetic progression. We will find the first and the last number between $10$ and $200$ that are divisible by $7$. Then we will find the total number of terms from the formula for the general term of an arithmetic progression. At last, we will use the formula for the sum of terms of an arithmetic progression with the first term and the last term known and also the number of terms known.

Complete answer:
The first number after $10$ that is divisible by $7$ is $14$. To find the last number under $200$ that will be divisible by $7$, we will divide $200$ by $7$, and then subtract the remainder from $200$.
When we divide $200$ by $7$ we get the remainder as four. So, the last number is $196$.
Now, we know that the nth term of an arithmetic progression is:
$n{\text{th term }} = a + \left( {n - 1} \right)d$

$$a = {\text{ the first term}} \\\ d = {\text{ the common difference}} \\\ n = {\text{ number of terms}} \\\$$

So, putting the values we get;
$14 + \left( {n - 1} \right)7 = 196$
On shifting the term to RHS and subtracting we get;
$\Rightarrow \left( {n - 1} \right)7 = 182$
On dividing we get;
$\Rightarrow \left( {n - 1} \right) = 26$
So, we get;
$\Rightarrow n = 27$
Now we know that the sum of n terms of an arithmetic progression is:
$S = \dfrac{n}{2}\left( {a + l} \right)$
$S = {\text{ sum of terms}}$
$l = {\text{ last term}}$
So, putting the values we get;
$\Rightarrow S = \dfrac{{27}}{2}\left( {14 + 196} \right)$
On adding we get;
$\Rightarrow S = \dfrac{{27}}{2} \times 210$
On solving we get;
$\Rightarrow S = 2835$

Therefore, the correct option is b

Note: If someone does not know the concept of arithmetic progression, he can think of writing all the numbers divisible by seven and then adding them up. In this case, this is possible also. But in some other questions, if the interval is very large then it will not be possible to list the numbers and add them up, so it is useful to use this concept.