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Find the sum of all odd integers between 2 and 100 which are divisible by 3.

Answer»

The odd integers between 2 and 100 which is divisible by 3 are 3, 9, 15, ..., 99. 

First and last odd integer between 2 and 100 which is divisible by 3 are 3 and 99 respectively. 

First term of given sequence is a = 3. 

And common difference of AP is d = a2 – a1 = 9 –3 = 6. 

Let 99 is nth term of sequence 3, 9, 15, .………., 99. 

∴ a + (n – 1) d = 99. (∵ an = a + (n – 1) d) 

⇒ 3 + (n – 1)6 = 99 (∵ a = 3 & d = 6) 

⇒ 6(n – 1) = 99 – 3 = 96 

⇒ n – 1 = \(\frac{96}{6}\) = 16 

⇒ n = 16 + 1 = 17. 

Now, sum of sequence 3, 9, 15, ………, 99 = \(\frac{n}{2}\)[a + an]

= \(\frac{17}{2}\) [3 + 99] (∵ n = 17, a = 3 & an = 99) 

= \(\frac{17}{2}\) × 102 = 17 × 51 = 867. 

Hence, the sum of all odd integers between 2 and 100 which are divisible by 3 is 867.


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