What is the sum of all numbers ending in 2 or 9 between 1000 and 2000?
Answer is 300 100
so you want the sum of
(1002+1012+1022+...+2992) + (1009+1019+...+1999)
two AP's , one with a = 1002, d=10, the other with a=1009, d=10
How many terms in each one?
First: term(n) = a + (n-1)d
2992 = 1002 + 10(n-1)
1990 = 10n - 10
n = 200
Find the number of terms in the second series (the number of terms is not 200).
Find the sum of each series using your favourite formula, then add the two sums.
Makes sense. Thank you
To find the sum of all numbers ending in 2 or 9 between 1000 and 2000, you can follow these steps:
1. Determine the first number ending in 2 or 9. In this case, it is 1002 (the smallest number between 1000 and 2000 that ends in 2).
2. Determine the last number ending in 2 or 9. In this case, it is 1992 (the largest number between 1000 and 2000 that ends in 2).
3. Calculate the total number of numbers between 1002 and 1992 (inclusive) that end in 2 or 9. To do this, subtract the first number (1002) from the last number (1992) and add 1 to include both endpoints: 1992 - 1002 + 1 = 991.
4. Calculate the sum of the first and last number: 1002 + 1992 = 2994.
5. Calculate the sum of all the numbers between 1002 and 1992 by multiplying the total count (991) by the average of the first and last number: 991 x (1002 + 1992) / 2 = 991 x 2994 / 2 = 1489007.
Therefore, the sum of all numbers ending in 2 or 9 between 1000 and 2000 is 1489007.