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What is the (n)th term of the sequence \(18,9,\frac{9}{2},\frac{9}{4},\ldots\)?श्रेणी \(18,9,\frac{9}{2},\frac{9}{4},\ldots\) का (n)वां पद क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(18\left(\frac{1}{2}\right)^{n-1}\)\(18\left(\frac{1}{2}\right)^{n-1}\)
Each term is halved, so \(a_n=18\left(\frac{1}{2}\right)^{n-1}\). In a geometric sequence, the exponent starts with (n-1).हर पद आधा हो रहा है इसलिए \(a_n=18\left(\frac{1}{2}\right)^{n-1}\)। गुणोत्तर श्रेणी में घात (n-1) से शुरू होती है।
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