Question

Consider the numbers between 100 and 300 which leave the reminder 2 on divided by 3
a) which is the first number of the sequence?
b)which is the last number of the sequence?
c) How many such numbers are in the sequence?
d)Find the sum of all numbers in the sequence ? ​

Answer 1

Numbers between 100 and 300 which leave remainder 2 on division by 3 are 101, 104, 107, 110, .........299

(a) First number of this sequence = 101

(b) Last number of this sequence = 299

(c) an=a+(n−1)d

299 = 101+(n−1)3

299-101 = (n-1) 3

198/3 = n-1

n = 66+1 = 67

Total numbers ∴n=67

(d) Sum of all numbers in the sequence = n/

2(a+an)

67/2 ( 101+299)

67×200

∴ Sum of all numbers = 13400