Solveeit Logo

Question

Question: The sum of integers from 1 to 100 which are divisible by 2 or 5 is- a) 300 b) 3050 c) 3200 d...

The sum of integers from 1 to 100 which are divisible by 2 or 5 is-
a) 300
b) 3050
c) 3200
d) 3250

Explanation

Solution

Hint: Here, in this given question, first of all we have to find the sum of all the integers from 1 and 100 which are divisible by 2 or 5 by using the concept of Arithmetic Progressions (AP). From the obtained sum, we have to subtract the sum of all those numbers which were added twice that are the numbers divisible by 2×5=102\times 5=10 as they were added both as the multiples of 2 as well as of 5. The result will give us our answer.

Complete step by step solution:
In this question, we are asked to find the sum of integers from 1 to 100 which are divisible by 2 or 5.
Now, we know that the sum of all the terms of an Arithmetic Progression is S=n2(a+an)S=\dfrac{n}{2}\left( a+{{a}_{n}} \right), where, n is the number of terms, a is the first term and an{{a}_{n}} is the last term of the series.
So, for the sum of all the integers divisible by 2 from 1 and 100, we can get the parameters as
a=2a=2 , an=100{{a}_{n}}=100, n=1002=50n=\dfrac{100}{2}=50
Now, substituting all these in formula for sum, we have
S1=502(2+100)=50×51=2550..............(1.1){{S}_{1}}=\dfrac{50}{2}\left( 2+100 \right)=50\times 51=2550..............(1.1)
For the sum of all the integers divisible by 5 from 1 and 100, we can get the parameters as
a=5a=5,an=100{{a}_{n}}=100, n=1005=20n=\dfrac{100}{5}=20
Let us substitute all these in formula for sum, then we have
S2=202(5+100)=10×105=1050..............(1.2){{S}_{2}}=\dfrac{20}{2}\left( 5+100 \right)=10\times 105=1050..............(1.2)
Now, we have to subtract the sum of the integers which are added twice that are the numbers divisible by2×5=102\times 5=10.
For the sum of all the integers divisible by 10 from 1 and 100, we can get the parameters as
a=10a=10, an=100{{a}_{n}}=100, n=10010=10n=\dfrac{100}{10}=10
Now, substituting all these in formula for sum, we have
S3=102(10+100)=10×55=550..............(1.3){{S}_{3}}=\dfrac{10}{2}\left( 10+100 \right)=10\times 55=550..............(1.3)
So, the sum of integers from 1 to 100 which are divisible by 2 or 5 is
S4=S1+S2S3 =2550+1050550 =3050..............(1.4) \begin{aligned} & {{S}_{4}}={{S}_{1}}+{{S}_{2}}-{{S}_{3}} \\\ & =2550+1050-550 \\\ & =3050..............(1.4) \\\ \end{aligned}
Hence, we have obtained the sum of integers from 1 to 100 which are divisible by 2 or 5 as 3050.
Therefore, the correct answer to the question is option (b) 3050.

Note: In this sort of question, we must be careful while choosing the first and the last terms of the Arithmetic Progressions (AP). If the between term is used in the question, the given first and last numbers should not be included in the analysis but as here, if form is given then both of the numbers need to be included in the analysis process.