### Consider an arithmetic series and a geometric series ....

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.