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 ____________.
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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The only AP is 11, 16, 21, 26 and the only GP is 4, 8, 16, 32; find the largest four-digit last terms and the common elements.
Updated On: Nov 4, 2026
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Correct Answer: 3
Solution and Explanation
AP: the four numbers 11, 16, 21, 26 (common difference 5). GP: the four numbers 4, 8, 16, 32 (common ratio 2).
AP terms are \(11 + 5(k-1)\), i.e. numbers \(\equiv 1 \pmod 5\) from 11 up to the largest four-digit such number, 9996.
GP terms are \(4\cdot 2^k\); the largest four-digit one is \(4 \cdot 2^{11} = 8192\) (k = 0 to 11).
Residues of \(4\cdot2^k\) mod 5 for k = 0..11: 4, 3, 1, 2, 4, 3, 1, 2, 4, 3, 1, 2. The residue is 1 for k = 2, 6, 10: terms 16, 256, 4096, all \(\ge 11\) and \(\le 9996\).
Common terms: 16, 256, 4096, i.e. 3.
Answer: 3
AP terms are \(11 + 5(k-1)\), i.e. numbers \(\equiv 1 \pmod 5\) from 11 up to the largest four-digit such number, 9996.
GP terms are \(4\cdot 2^k\); the largest four-digit one is \(4 \cdot 2^{11} = 8192\) (k = 0 to 11).
Residues of \(4\cdot2^k\) mod 5 for k = 0..11: 4, 3, 1, 2, 4, 3, 1, 2, 4, 3, 1, 2. The residue is 1 for k = 2, 6, 10: terms 16, 256, 4096, all \(\ge 11\) and \(\le 9996\).
Common terms: 16, 256, 4096, i.e. 3.
Answer: 3
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