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What is the (n)th term of the sequence \(125,25,5,1,\frac{1}{5},\ldots\)?श्रेणी \(125,25,5,1,\frac{1}{5},\ldots\) का (n)वां पद क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(125\left(\frac{1}{5}\right)^{n-1}\)\(125\left(\frac{1}{5}\right)^{n-1}\)
Each term is multiplied by \(\frac{1}{5}\), so \(a_n=125\left(\frac{1}{5}\right)^{n-1}\). Keep the first term fixed and use exponent (n-1).हर पद \(\frac{1}{5}\) गुना हो रहा है इसलिए \(a_n=125\left(\frac{1}{5}\right)^{n-1}\)। पहले पद को सुरक्षित रखकर (n-1) घात लगाएं।
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