Question

how many 3 digit no. are there between 100 and 550 110 and 550 are not inculded ​

Answer 1

Answer:

The fastest way I can think of to solve this is to sum up ALL integers between 100 and 550, and subtract those divisible by 9. So:

Σ(100–550)n=Σ(1–550)n - Σ(1–99)n

=550(551)/2 - 99(100)/2

=146575

The first number in between 100 and 550 divisible by 9 is 108, or 12 x 9. The last is 549, or 61 x 9. So, we add up the sum of 9n, from 12 to 61:

Σ(12–61)9n=Σ(1–61)9n - Σ(1–11) 9n

=9Σ(1–61)n - 9Σ(1–11)n

=9[(61)(62)/2 - (11)(12)/2]

=9[1891–66]

=16425

Then, the sum of all integers NOT divisible by 9, between 100 and 550 is:

T=146575–16425

=130150

Step-by-step explanation:

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Answer 2

Answer:

hey your answer is as follows

so pls mark it as brainliest

Step-by-step explanation:

now basically in this sum we need to apply concept of arithmetic progression ie ap

so the three digit numbers between 100 and 550 except 100 and 550 are 101,102,103,so on upto 549

thus the common difference d between each of the numbers is constant ie 1

so as the common difference is constant

the above series of numbers form a sequence of arithmetic progression ie ap

so for the above ap

a=t1=first term=101

tn=last term=549

d=1

we need to find n

so using

tn=a+(n-1)d

ie 549=101+[(n-1)×1]

so 549=101+n-1

ie 549=100+n

so n=449

hence in total there are 449 three digit numbers between 100 and 550 except 100 and 550