Answer

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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.

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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