Consider the sequence of two digit numbers which leave a remainder 1 on divisible
by 5 .
a ) What is its common difference ?
b) Which is the smallest and largest numbers in this sequence ?
c ) What is algebraic form of this sequence ?
d ) How many two digit numbers are there which leave a remainder 1 on divisible by 5 ?
e) What is the sum of such numbers ?
Answers
Given : sequence of two digit numbers which leave a remainder 1 on division by 5
To Find : the sequence .
Solution:
leave a remainder 1 on division by 5
can be represented by 5n+ 1
as sequence of two digit numbers
Hence 10 ≤ 5n + 1 ≤ 99
10 ≤ 5n + 1
=> 9 ≤ 5n
=> 2 ≤ n
5n + 1 ≤ 99
=> n ≤ 19
2 ≤ n ≤ 19
Hence sequence is
11 , 16 , 21 , ___________________ 91 , 96
5n + 1 , 2 ≤ n ≤ 19
common difference is 5
smallest and largest numbers in this sequence are 11 and 96 respectively
algebraic form of this sequence 5n + 1 , 2 ≤ n ≤ 19
18 such numbers are there
Sum = (18/2) ( 11 + 96) = 963
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