Advertisements
Advertisements
प्रश्न
How many two-digits numbers are divisible by 3?
उत्तर
The two-digit numbers divisible by 3 are 12, 15, 18, ..., 99.
Clearly, these number are in AP.
Here, a = 12 and d =15 - 12 = 3
Let this AP contains n terms. Then,
an = 99
⇒ 12 + (n-1) × 3 =99 [an = a + (n-1) d]
⇒ 3n + 9 =99
⇒ 3n = 99 - 9 = 90
⇒ n = 30
Hence, there are 30 two-digit numbers divisible by 3.
संबंधित प्रश्न
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]
Find the sum of first n odd natural numbers
If the 8th term of an A.P. is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of A.P.
Which term of the AP 21, 18, 15, … is zero?
Find the 25th term of the AP \[- 5, \frac{- 5}{2}, 0, \frac{5}{2}, . . .\]
The sum of third and seventh term of an A. P. is 6 and their product is 8. Find the first term and the common difference of the A. P.
The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.
Write the formula of the sum of first n terms for an A.P.
If ₹ 3900 will have to be repaid in 12 monthly instalments such that each instalment being more than the preceding one by ₹ 10, then find the amount of the first and last instalment
Find the sum of first 16 terms of the A.P. whose nth term is given by an = 5n – 3.
