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 _______.

Step-by-step solution

Given set = 11 , 8 , 21 , 16 , 26 , 32 , 4 Given an arithmetic series and a geometric series having the first four terms from the given set. Then, Geometric series   :   4 , 8 , 16 , 32 , 64 , 128 , 256 , 512 , 1024 , 2048 , 4096 , 8192 Arithmetic series :   11 , 16 , 21 , 26 , 31 , 36 , . . . , 256 , . . . . 4096 , . . . T n   ( G . P . ) = 4 · 2 n - 1 T n   ( A . P . ) = 11 + 5 n - 1 = 5 n + 6 For common terms, 4 · 2 n - 1 = 5 n + 6 So, common terms are 16 , 256 , 4096