1st AP=17,21,25.....417 the general nth term, t=17+(n-1)4
2nd AP=16,21,26....466 the general mth term for this AP p=16+(m-1)5
for common terms in both APs,we solve for the general terms by making them equal as.. t=p or 17+(n-1)4=16+(m-1)5 5m-4n=2 or m=(4n+2)/5 now putting the values of n as 1,2,3,4.....101(the last term) nd getting the integral values of m like we get m=2nd term by putting n=2...
4 comments:
is the ans=20
yeah correct...plz do explain..
1st AP=17,21,25.....417
the general nth term,
t=17+(n-1)4
2nd AP=16,21,26....466
the general mth term for this AP
p=16+(m-1)5
for common terms in both APs,we solve for the general terms by making them equal as..
t=p or
17+(n-1)4=16+(m-1)5
5m-4n=2 or
m=(4n+2)/5
now putting the values of n as 1,2,3,4.....101(the last term) nd getting the integral values of m like we get m=2nd term by putting n=2...
solution :
common diff. of 1st AP,D1=4
D2=5
now lcm of d1 and d2 is will be the c.d. of common AP. i.e D(common)=20
now 1st common term=21,d-20..next term=41...
and for last term
417>= a+(n-1)d
417>= 21 +20d -20
on solving...
416/20=20.8
therefore we'll b considering 20 terms as the next term would be greater dan 417(of the first ap)
thrfr ans z 20
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