5 votes 5 votes How many A.P's with $10$ terms are there whose first term is in the set$\{1,2,3\}$ and whose common difference is in the set $\{1,2,3,4,5\}$ ? Quantitative Aptitude combinatory + – Subarna Das asked Jan 23, 2018 edited Mar 11, 2018 by Sukanya Das Subarna Das 551 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
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. Sukanya Das answered Jan 23, 2018 selected Jan 25, 2018 by Subarna Das Sukanya Das comment Share Follow See all 0 reply Please log in or register to add a comment.
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. Avdhesh Singh Rana answered Jan 23, 2018 Avdhesh Singh Rana comment Share Follow See all 0 reply Please log in or register to add a comment.
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 Rishi yadav answered Jan 24, 2018 Rishi yadav comment Share Follow See all 0 reply Please log in or register to add a comment.