Math, asked by amishisonline, 1 month ago

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?​

Answers

Answered by leondalmeida
0

Answer:

Step-by-step explanation:

a) 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) by the following formula:

Tn = a +(n-1)d\\997 = 103 + (n-1)3\\997 = 103 + 3n - 3\\997 - 100 = 3\\3n = 897\\n = 299

SO, THERE ARE 299 SUCH NUMBERS

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