Question

find the sum of all natural numbers between 100 and 250 which leaves a remainder 3 when divided by 6

Answer 1

Answer:

4425

Step-by-step explanation:

The numbers are 105, 111, 117,........., 249.

We use formula of nth term of an AP(arithmetic progression).

249 = 105 + (n-1)6

144/6 = n-1

24+1 = n

n = 25

Now we use the formula of sum of n terms of AP

S = 25/2(2x105 + (25-1)6)

S = 25/2(210 +144)

S = 25/2(354)

S = 25x177

S = 4425 answer

Answer 2

The natural number which leaves remainder 2 greater than 100 is 102, and less than 200 is 197.

The series is 102, 107, 112, 117, 122, .... , 197.

Number of terms

=

5

200

−

5

100

=40−20=20

S

n

=

2

n

[a+l]=

2

20

[102+197]=2990