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# Find the sum of odd integers between 2 and 100 divisible by 3.

Last updated date: 18th Mar 2023
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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 \\ \$