JEE Main 16 March 2021 Shift 1 Solved Paper

Given, set {11,8,21,16,26,32,4}
By observation, we can say that
AP={11,16,21,26,...}
GP={4,8,16,32,...}
Since we are looking for common terms in both the series where 5m+6 is m‌th ‌ term of AP and 2n+1 is n‌th ‌ term of GP
5m+6=4⋅2n−1
5m+6=2n+1
So, (2n+1−6) should be a multiple of 5 .
The unit digit of 2k is 2,4,6,8.
So, when 6 is subtracted from 2n+1, the possible unit digits will be 6 , 8,0,2.
Only 0 is divisible by 5 .
Hence, 2n+1 unit digit has to be 6 .
2n+1=24,28,212,216...
As, 216 will not be a 4 digit number, so, common terms ={16,256,4096}
∴ Number of common terms =3