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

Answer

Verified

364.5k+ views

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.

Last updated date: 28th Sep 2023

•

Total views: 364.5k

•

Views today: 5.64k

Recently Updated Pages

What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE