edited by
551 views

3 Answers

Best answer
3 votes
3 votes

The first term can be 1/2/3 .

So, the first term can be chosen in 3 ways.

And for each first term we can choose common difference in 5 ways from the common difference set (1/2/3/4/5)

So, total ways = 3 * 5 ways

                       = 15 ways.

selected by
0 votes
0 votes
There 3 ways to choose the first term of an AP and there are 5 ways to choose common difference for an AP.
Total 3 x 5 = 15 ways are there.
0 votes
0 votes
First term is in the set {1,2,3} so we can select one element so by 3C1 ways we can select first similiarly we can select difference from second set so 5C1 hence no of APs = 5C1*3C1 = 15

Related questions

1 votes
1 votes
2 answers
1
Pooja Khatri asked May 16, 2019
590 views
How many 4 letter combinations can be made with the help of letters of the word STATISTICS?
0 votes
0 votes
0 answers
2
Ayush Upadhyaya asked Oct 27, 2018
384 views
If $^nP_6=332640$How to quickly and efficiently find out the value of n.I know that fact that product of m consecutive integers is divisible by m!.How can we find the val...
1 votes
1 votes
2 answers
3
imnitish asked Jul 18, 2018
374 views
How many $10$ - digit strings of $0's$ and $1's$ are there that do not contain any consecutive $0's$?
3 votes
3 votes
1 answer
4
Subarna Das asked Jan 23, 2018
505 views
How many natural number not exceeding 4321 can be formed with the digits 1, 2, 3, 4, if the digits can be repeated?