Question

Find the sum of natural numbers between 100 to 300 divisible by 3​

Answer 1

Answer:

6633

Step-by-step explanation:

We have to find the sum of the numbers b/w 100-300 which is divisible by 3

so the sequence be look like

102,105,108,...….,300

By seeing above series we got that it is in AP

by observing above series

T n=300

a=102

d=3

we know that

T n=a+(n-1)*d

substituting the known data

300=102+(n-1)*3

198=(n-1)*3

(n-1)=66

n=67

we know that

Sn=(n/2)*(2a+(n-1)*d)

   =(67/2)(2*102+(66)*3)

   =6633

so the sum of natural numbers between 100 to 300 divisible by 3 is given by 6633

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