Question

find the sum of numbers which are divisible by 2 or 3 between 1 and 100(including 100) by sequence and series

Answer 1

Answer:-

• The sequence of numbers which are divisible by 2 between 1 and 100 is 2 , 4 , 6 ... 100.

• The sequence of numbers which are divisible by 3 between 1 and 100 is 3 , 6 , 9 , ... 99.

Common terms from both the sequences are 6 , 12, 18 ,...96.

If we assume that this sequence is in AP,

• a(first term) = 6
• d(common difference) = 12 - 6 = 6
• nth term = 96.

We know that,

nth term of an AP$\sf{ (a_n)}$ = a + (n - 1)d.

→ 6 + (n - 1)(6) = 96

→ 6 + 6n - 6 = 96

→ 6n = 96

→ n = 96/6

→ n = 16

We know that,

Sum of first n terms of an AP$\sf (S_n)$ = n/2 [2a + (n - 1)d]

→ $\sf{{S}_{16} = \frac{16}{2} [2(6) + (16 - 1)(6)]}$

→ $\sf{ {S}_{16} = 8(12 + 90)}$

→ $\sf{{S}_{16}= 8(102)}$

→ $\sf\large{{S}_{16 }= 816}$

Hence, the sum of numbers which are divisible by both 2 & 3 is 816.