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the first erm of a geometric series is 3, and the sum of the first term and the second term is 15. what is the sum of the first six term? A.1023 B.3906 C.4096 D.11718
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i know that the second term is 12
the common ratio r= 2nd term / 1st term = 12/3 = ... ? now just use the formula for sum, with a1 = 3, and n = 6 \(\large S_n=a_1\dfrac{r^n-1}{r-1}\)
i dont know what to plug in to what
you could get r= 12/3 = 4, right ? plug in r= 4,a1 = 3 and n=6
\[3\frac{ 95 }{3 }\]
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