Question

9) Find the sum of all numbers between 250 and 1000 which are exactly divisible by 3.​

Answer 1

Answer:

The numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, ......,999

This is an A.P. whose first term a = 252, Common difference, d = 3 and Last term l=999

We Know that

l=a+(n-1)d

999=252+(n-1)3

999 - 252 = 3(n - 1)

n=250

We know that

S=n/2(a+l)

=250/2(252+999)

=125*1251

=156375

Hence, the required sum is 156375.(Ans)

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Answer 2

Answer:

Numbers between 250 and 1000 divisible by 3 = 252, 255, 258,......... 999.

and, Common difference (d) = 255-252

                                                 = 3.

Therefore, a = 252

                  l = 999

                  d = 3

As we know:  l = a + (n-1)d

             999 = 252 + (n-1)3

             999 = 252 + 3n - 3

             999 = 3n + 249

             999 - 249 = 3n

             750 = 3n

             750/3 = n

             250 = n

Therefore, the total no. of terms b/w 250 and 1000 divisible by 3 = 250

 Now:

           Sum of all terms = n/2 [2a + (n-1)d]

                                       = 250/2 [2 X 252 + (250-1)3]

                                       = 125 [504 + 747]

                                       = 125 X 1251

                                       =  156375

Step-by-step explanation:

Already well explained above :)

 Hope it helps