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
Solution:
GP : $4,8,16,32,64,128,256,512,1024,2048$
4096,8192
$\mathrm{AP}: 11,16,21,26,31,36$
Common terms: $16,256,4096$ only