Question
Download Solution PDFGP मध्ये 4 पदांची बेरीज शोधा, दिलेली पहिली संज्ञा 15 आहे आणि सामान्य गुणोत्तर 4 आहे.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिले:
GP ची पहिली टर्म = 15
GP चे सामान्य गुणोत्तर = 4
वापरलेले सूत्र:
\(S_n=a_1\times \dfrac{(r^n-1)}{(r-1)}\)
गणना:
\(S_4=15\times \dfrac{(4^4-1)}{(4-1)}\) \(= \dfrac{15\times 255}{3} = 1275\)
∴ उत्तर 1275 आहे.
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