If 2,5,8... upto 50 terms--- (1) and 3,5,7,... upto 60 terms--- (2) terms are two APs then nth term in (1) will be a+(n-1)d --- (3) and in (2) kth term will be a+(k-1)d --- (4). Now on equating (3) and (4) i got 3n=2(k+1).
Please help me in what to do ahead of this and please also explain each step.
Difference in these sequences are 3 and 2 therefore after the first common term (a) every term of the form a+6b will be in both sequences. The first common one is 5. We also need to look at which sequence terminates first. It is the second one with 121 as its last term. The last common term will be the biggest number of the form 5+6b is 119, which is equal to 5+19*6. Since 5= 5+0*19 we conclude that there are 20 common terms.