Question

Consider the sequence of two digit natural numbers which leave remainder 1 on divided by 5. a) What is its common difference? b) What is the smallest number in this sequence? c) What is the largest number in this sequence? d) How many numbers are there in this sequence?​

Answer 1

Answer:

a) Let the first term be a and the common difference be d.

So, the series will be a, a+d, a+2d

If we subtract 1 from each number in the sequence, it will be divisible by 3 and will look like( a-1,a+d-1,a+2d-1), now their common difference will be divisible by 3

Such as,

(a+d-1) -(a-1) = 3n

a+d-1-a+1=3n

d=3n

So,common difference will be 3n where n is a Natural number

b) Smallest 3 digit number divisible by 3 = 102, so the smallest number in this sequence is 102+1=103

c) The largest 3 digit number that leaves remainder 1 when divided by 3 = 997

We have smallest number as 103 and largest as 997,if we put all such numbers in sequence, we will get 3 as a common difference, for example,

103,106,109...997

So, by the following formula:

SO, THERE ARE 299 SUCH NUMBERS

Step-by-step explanation:

hope this helps.