Question

Find the sum of all odd integers between 1 and 100 which are divisible by 3

Answer 1

according to the question , 3+9+15+21…..99

The common thing here is,

3(1+3+5+7….33)

Now the summation of ’n’ odd number is n^2 ( where n is the number of terms)

Here, number of terms is 17

So sum of first seventeen odd natural number is n^2 = 17^2 = 289

Now coming to the main question,

Sum of all odd integers between 2 and 100 which is divisible by 3 is 3 (289) which is equal to 867.

Answer 2

Hi there!
Here's the answer:

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¶¶¶ POINTS TO REMEMBER :

• Sum of 'n' Odd No.s starting from 1= $n^{2}$

• 1 + 3 + 5 + ………+ k ; where k is last odd no. in the series

Then No. of Odd No.s in the series 'n' :
$n = \frac{33 +1}{2}$

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SOLUTION:

• Odd Integers between 1 - 100 that are divisible by 3 are the ODD MULTIPLES OF 3 FROM 1 - 100

viz ., 3, 9, 15, 21, ……… , 99
= 3(1), 3(3), 3(5), 3(7), ……… , 3(33)

The Given odd Integers are represented as 3i ; where i = {1, 3, 5, 7, … 33}

The sum of these No.s
= 3 + 6 + 9 + 12 + …… + 99
= 3(1 + 3 + 5 + …… + 33)

Here we use formula,
last term k = 33
Now, find 'n'

=> $n = \frac{33 + 1}{2} = \frac{34}{2} = 17$

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$Sum = 3(17^{2}) = 3 × 289 = 867$

$\underline{\underline{Required\: Sum= 867}}$

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:)

Hope it helps