If ` T₂ + T₅ = 8`of an A.P & `T₃ + T₇ = 14`of that A.P then, find the 11th term?

Explanation

Let F, H and B be the sets B of television watchers who watch Football, Hockey and Basketball, respectively. Then, according to the problem, we have
`n(U)=500,n(F)=285,n(H)=195`,
`n(B)=115,n(FnnB)=45`,
`n(FnnH)=70,n(HnnB)=50`
and `n(F'uuH'uuB')=50`,
where U is the set of all the television watchers.
Since, `n(F'uuH'uuB')=n(U)-n(FuuHuuB)`
`implies 50=500-n(FuuHuuB)`
`implies n(FuuHuuB)=450`
We know that,
`n(FuuHuuB)=n(F)+n(H)+n(B)-n(FnnH)-n(HnnB)-n(BnnF)+n(FnnHnnB)`
`implies450=285+195+115-70-50-45+n(FnnHnnB)`
`therefore n(FnnHnnB)=20`
which is the number of those who watch all the three games. Also, number of persons who watch football only `= n(FnnH'nnB')`
`=n(F)-n(FnnH)-n(FnnB)+n(FnnHnnB)`
`=285-70-45+20=190`
The number of persons who watch hockey only
`=n(HnnF'nnB')`
`=n(H)-n(HnnF)-n(HnnB)+n(HnnFnnB)`
`=195-70-50+20=95`
and the number of persons who watch basketball only
`=n(BnnH'nnF')`
`=n(B)-n(BnnH)-n(BnnF)+n(HnnFnnB)`
`=115-50-45+20=40`
Hence, required number of those who watch exactly one of the three games
= 190 + 95 + 40 = 325