Question

find out the sum of all natural numbers between 1 and 145 which are dividible by 4

Answer 1

Solution:

The given problem can be solved using the knowledge of Arithmetic Progression.

$\text{Sum on 'n' terms of an AP} = \dfrac{n}{2}\:\text{[ 2a + ( n - 1 ) d ]}$

Here,

• 'a' refers to the first term
• 'd' refers to the common difference
• 'n' refers to the number of terms

Finding the Arithmetic Progression: Since it is given that numbers in the progression are divisible by 4 and the numbers lie between 1 to 145, the first number of the progression would be 4. Now since the terms are a multiple of 4, the common difference would be 4.

Hence the AP is 4, 8, 12, ..., 144

Here,

• a = 4
• d = 4
• n = ?
• l = 144

Formula to calculate 'n':

Last Term ( l ) = a + ( n - 1 ) d

Substituting the values we get,

⇒ 144 = 4 + ( n - 1 ) 4

⇒ 144 - 4 = ( n - 1 ) 4

⇒ 140 = ( n - 1 ) 4

⇒ 140 / 4 = n - 1

⇒ 35 = n - 1

⇒ n = 35 + 1 = 36

Hence the number of terms in the given AP is 36.

Now substituting the values in the Sum of 'n' terms Formula we get,

$\implies S_n = \dfrac{36}{2}\:\text{[ 8 + ( 36 - 1 ) 4 ]}\\\\\implies S_n = 18 \: \text{[ 8 + ( 35 ) 4 ] }\\\\\implies S_n = 18 \: \text{[ 8 + 140 ]}\\\\\implies S_n = 18 \times 148 = \boxed{2664}$

Hence the Sum of all natural numbers between 1 and 145 which are divisible by 4 is 2664.