Saved Bookmarks
| 1. |
Consider the sequence of two digit numbers which leave a remainder 1 on divisibleby 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 ? |
|
Answer» 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 Learn More: In an A.P if sum of its first n terms is 3n square +5n and it's KTH term ... Find the sum of frist 51 terms of the AP whose 2nd term is 2 and 4th ... |
|