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# Find the sum of odd integers between 2 and 100 divisible by 3.  Answer Verified
Hint: First try to analyse the given scenario and then form an AP series, this will make the simplification easy.
According to the question, the numbers will be $3,6,9,12........,99$ .The first three terms are $3,6{\text{ and 9}}$ . Let’s compute ${2^{nd}} - {1^{st}}{\text{ term}}$ and ${3^{rd}} - {2^{nd}}{\text{ term}}$ . This will give us $6 - 3 = 3$ and $9 - 6 = 3$ .Since these differences are equal so we’ll say the given series is an AP. First term (a) is 3, last term $(l)$ is 99 and the common difference, which is the difference between two consecutive terms, is 3.
Now, we want to know which number term is 99. We’ll use ${n^{th}}$ term $({T_n})$ formula, which is $a + (n - 1)d$ . Where, a is the first term of the series, n is the number of terms and d is a common difference.

$\ {T_n} = a + (n - 1)d \\ \Rightarrow 99 = 3 + (n - 1)3 \\ \Rightarrow 99 - 3 = (n - 1)3 \\ \Rightarrow 96 = (n - 1)3 \\ \Rightarrow \dfrac{{96}}{3} = n - 1 \\ \Rightarrow 32 = n - 1 \\ \Rightarrow 32 + 1 = n \\ \Rightarrow n = 33 \\ \$

So, ${T_{33}}$ is $99$ . Now, we’ll use the sum formula of AP which is ${S_n} = \dfrac{n}{2}(a + l)$ .On putting the values in this formula,

$\ {S_{33}} = \dfrac{{33}}{2}(3 + 99) \\ \Rightarrow {S_{33}} = \dfrac{{33}}{2}(102) \\ \Rightarrow {S_{33}} = 33 \times 51 \\ \Rightarrow {S_{33}} = 1683 \\ \$
Hence the required sum is 1683.
Note: The hack in this question is to recognise the series. Once you are done with it, just applying the formula will give us the answer.
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Sum of Odd Numbers  Sum of Odd Numbers Formula  Geometric Progression Sum of GP  Multiplication and Division of Integers  Even and Odd Numbers  Even and Odd Functions  Multiplication and Division of Integers - Rules  Arithmetic Progression  Properties of Integers  Operations of Integers  