Question

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 ___

Answer 1

Finding the terms:

A.P : $11,16,21,26\dots \dots .$

G.P : $4,8,16,32\dots \dots .$

Extending A.P. and G.P.

A.P : $11,16,21,26,31,.....,256,261,....,4096,....$

G.P : $4,8,16,32,64,128,256,512,1024,2048,4096,8192,....$

So common terms are $16,256,4096$.

Hence, the number of common terms in these two series is equal to _3___