Consider an arithmetic series and a geometric series having four initial terms from the set {11,8,21,16,26,32,4}. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to _ _ _ _ .
Solution
Given set $ \equiv $ {11,8,21,16,26,32,4} $ \equiv $ {4,8,11,16,21,26,32}
GP's 1st 4 terms $\equiv $ 4, 8, 16, 32
AP's 1st 4 terms $\equiv $ 11, 16, 21, 26
The whole GP $\equiv $ 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192
AP has either 1 or 6 digit at unit's place.
The terms of GP having 1 or 6 at unit's place $\equiv $ 16, 256, 4096
16 is clearly present in AP.
If 256 is a term of AP, then 256 = 11 + (n - 1 ) . 5 for $n \in N$ which is true.
If 4096 is a term of AP, then 4096 = 11 + (n - 1) . 5 for $n \in N $ which is also true.
Hence, 3 terms.
