Advertisements
Advertisements
Question
How many three-digit natural numbers are divisible by 9?
Solution
The three-digit natural numbers divisible by 9 are 108, 117, 126 ,….. 999.
Clearly, these number are in AP.
Here. a = 108 and d = 117 – 108 = 9
Let this AP contains n terms. Then,
an = 999
⇒ 108 +(n-1) × 9 = 999 [an = a+(n-1) d]
⇒ 9n + 99 = 999
⇒ 9n = 999- 99 =900
⇒ n = 100
Hence, there are 100 three-digit numbers divisible by 9.
APPEARS IN
RELATED QUESTIONS
The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Find the sum of all integers between 50 and 500, which are divisible by 7.
Find the sum 25 + 28 + 31 + ….. + 100
Which term of the AP 21, 18, 15, … is zero?
There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.
Suppose the angles of a triangle are (a − d), a , (a + d) such that , (a + d) >a > (a − d).
Find the sum of first 10 terms of the A.P.
4 + 6 + 8 + .............
The sum of first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is `(1)/(3)`. Calculate the first and the thirteenth term.
Find the sum of first 17 terms of an AP whose 4th and 9th terms are –15 and –30 respectively.
Find the middle term of the AP. 95, 86, 77, ........, – 247.