Consider an arithmetic series and a geometric | vedclass

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(A) The geometric series $(GP)$ must start with $4$ and have a common ratio $r = 2$,so the terms are $4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 409

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$3$

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(A) The geometric series $(GP)$ must start with $4$ and have a common ratio $r = 2$,so the terms are $4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$. The next term $16384$ is a five-digit number.The arithmetic series $(AP)$ must start with $11$ and have a common difference $d = 5$ (since $16-11=5, 21-16=5, 26-21=5$),so the terms are $11, 16, 21, 26, 31, \dots, a_n = 11 + (n-1)5$.We look for terms in the $GP$ that are also in the $AP$. $A$ term $x$ is in the $AP$ if $x \equiv 11 \pmod{5}$,which is equivalent to $x \equiv 1 \pmod{5}$.Checking the $GP$ terms:$4 \equiv 4 \pmod{5}$$8 \equiv 3 \pmod{5}$$16 \equiv 1 \pmod{5}$ (Common)$32 \equiv 2 \pmod{5}$$64 \equiv 4 \pmod{5}$$128 \equiv 3 \pmod{5}$$256 \equiv 1 \pmod{5}$ (Common)$512 \equiv 2 \pmod{5}$$1024 \equiv 4 \pmod{5}$$2048 \equiv 3 \pmod{5}$$4096 \equiv 1 \pmod{5}$ (Common)$8192 \equiv 2 \pmod{5}$The common terms are $16, 256, 4096$. There are $3$ such terms.

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 .......

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