Question
Download Solution PDF82,92,102,...,152 का समांतर माध्य क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFगणना:
दिया गया है,
अनुक्रम है \(8^{2},\,9^{2},\,10^{2},\,\dots,\,15^{2}\)
पदों की संख्या, \(n = 15 - 8 + 1 = 8\)
प्रथम \(m\) वर्गों का योग है, \( \displaystyle \sum_{k=1}^{m} k^{2} = \frac{m(m+1)(2m+1)}{6}\)
\(15^{2}\) तक योग:
\(S_{15} = \frac{15 \times 16 \times 31}{6} = 1240\)
\(7^{2}\) तक योग:
\(S_{7} = \frac{7 \times 8 \times 15}{6} = 140\)
अभीष्ट पदों का योग:
\(S = S_{15} - S_{7} = 1240 - 140 = 1100\)
समांतर माध्य है,
\( \displaystyle \text{Mean} = \frac{S}{n} = \frac{1100}{8} = 137.5\)
∴ समांतर माध्य \(137.5\) है।
इसलिए, सही उत्तर विकल्प 3 है।
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