Question

Calculate the sum of numbers between $100{\text{ and }}150$ which are divisible by $13$.

A. $494$
B. $410$
C. $420$
D. $430$

Answer 1

Hint- In order to solve this problem first we will form an AP between $100{\text{ and }}150$ which are divisible by $13$, further we will find common difference and first term and at last by using the formula of sum of nth series we will get the required answer.

Complete step-by-step answer:

The numbers between $100{\text{ and }}150$ which are divisible by $13$are $104,117,.................143$

So AP = $104,117,.................143$\[2\]

Here first term is $104$

and common difference = second term \[ - \] first term \[ = {\text{ }}117 - 104{\text{ }} = {\text{ }}13\]

As we know that

The sum of first \[n\]terms of arithmetic series formula can be written as

\[{S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] \to (1)\]

Where \[n\]= number of terms = ?
\[a = \] first odd number \[ = 104\]
\[d = \]common difference of A.P. \[ = 13\]
\[{T_n} = \]Last term \[ = 143\]

So first we have to calculate total number of terms

As we know that \[{n^{th}}\] term of AP is given as

\[{T_n} = a + (n - 1)d\]

Substitute the values in above formula we have

\[
  143 = 104 + (n - 1)13 \\
  143 = 104 + 13n - 13 \\
  143 - 104 + 13 = 13n \\
  52 = 13n \\
  n = 4 \\
\]

Therefore total number of terms \[ = 4\]

Now put the values of \[n,a,d\] and \[{T_n}\] in formula of sum of \[n\] term of series

\[
  {S_4} = \dfrac{4}{2}\left[ {2 \times 104 + (4 - 1)13} \right] \\
   = 2[208 + (3)13] \\
   = 2[208 + 39] \\
   = 2[247] \\
   = 494 \\
\]

Hence, the sum of numbers between $100{\text{ and }}150$ which are divisible by \[13\] is \[494\] and the correct answer is option A.


Note- The formula states that the sum of our arithmetic sequence's first \[n\] terms is equal to \[n\] divided by \[2\] times the sum of twice the beginning term, \[a\], and the product of d, the common difference, and \[n\] minus \[1\]. The n represents the number of words we put together.