Question

301,308,315,…….. is an arithmetic sequence.

a) Write the remainder when the terms of this sequence are divided by 7?

b) Find the sum of all natural numbers between 300 and 500 that leave reminder two on

division by 7?

c) What is the sum of all natural numbers between 300 and 500 that leave reminder 3

on division by 7?​

Answer 1

Answer:

yes AP

a) Remainder is 0

b)303,310,317.......485,492,499

no. of terms=n

499=303+(n-1)7

(n-1)=196

n=29

sum=n/2(a+l)

29/2(303+499)

29/2*802

29*401

=11629

c)304,311,318,......486,493

493=304+(n-1)7

n=189/7+1

n=28

sum=28/2(304+493)

sum=797*14

=11158