Question
Download Solution PDFGP मधील 4 पदांची बेरीज शोधा, पहिली संज्ञा 16 आहे आणि सामान्य गुणोत्तर 6 आहे.
Answer (Detailed Solution Below)
Detailed Solution
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GP मधील 4 पदांची बेरीज, दिलेली पहिली टर्म 16 आहे आणि सामान्य गुणोत्तर 6 आहे.
वापरलेले सूत्र:
फॉर्मची प्रगती a, ar, ar 2 , ar 3 , ........ आहे भौमितिक प्रगती म्हणून ओळखली जाते (जेथे, a प्रथम पद आहे आणि r हे सामान्य गुणोत्तर आहे)
GP च्या n पदांची बेरीज, Sn = \(\dfrac{a(r^n-1)}{(r-1)}\) , जेथे r > 1
गणना:
आता, a = 16, r = 6, n = 4
Sn = \(\dfrac{16 × (6^4 - 1)}{(6 - 1)}\)
⇒ \(\dfrac{16 × (1296 - 1)}{(6 - 1)}\)
⇒ \(\dfrac{16 × 1295}{5}\) = 16 x 259 = 4144
Sn = 4144
∴ उत्तर 4144 आहे
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