document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209805>Question 1209805</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let a_1 + a_2 + a_3 + dotsb be an infinite geometric series with positive terms. If a_2 = 10, then find the smallest possible value of<BR>\n" );
document.write( "a_1 + a_2 + a_3.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850591 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850591\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's solve this problem step-by-step.\r<BR>\n" );
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document.write( "Understanding Geometric Series\r<BR>\n" );
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document.write( "A geometric series has the form: a, ar, ar², ar³, ...<BR>\n" );
document.write( "a is the first term (a_1)<BR>\n" );
document.write( "r is the common ratio<BR>\n" );
document.write( "a_n = ar^(n-1)<BR>\n" );
document.write( "Given Information\r<BR>\n" );
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document.write( "The series has positive terms, so a > 0 and r > 0.<BR>\n" );
document.write( "a_2 = 10<BR>\n" );
document.write( "Finding a_1 and r\r<BR>\n" );
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document.write( "a_2 = ar^(2-1) = ar<BR>\n" );
document.write( "ar = 10<BR>\n" );
document.write( "a = 10/r<BR>\n" );
document.write( "Finding a_1 + a_2 + a_3\r<BR>\n" );
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document.write( "a_1 + a_2 + a_3 = a + ar + ar²<BR>\n" );
document.write( "Substitute a = 10/r:<BR>\n" );
document.write( "(10/r) + 10 + 10r<BR>\n" );
document.write( "Minimizing a_1 + a_2 + a_3\r<BR>\n" );
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document.write( "Let S = (10/r) + 10 + 10r. We want to minimize S.\r<BR>\n" );
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document.write( "Take the derivative with respect to r:\r<BR>\n" );
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document.write( "dS/dr = -10/r² + 10<BR>\n" );
document.write( "Set the derivative to zero and solve for r:\r<BR>\n" );
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document.write( "-10/r² + 10 = 0<BR>\n" );
document.write( "10 = 10/r²<BR>\n" );
document.write( "r² = 1<BR>\n" );
document.write( "r = ±1<BR>\n" );
document.write( "Consider positive r:\r<BR>\n" );
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document.write( "Since the terms are positive, r must be positive. Therefore, r = 1.<BR>\n" );
document.write( "Check the second derivative:\r<BR>\n" );
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document.write( "d²S/dr² = 20/r³<BR>\n" );
document.write( "When r = 1, d²S/dr² = 20 > 0, which means we have a minimum.<BR>\n" );
document.write( "Find a_1:\r<BR>\n" );
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document.write( "a = 10/r = 10/1 = 10<BR>\n" );
document.write( "Find a_1 + a_2 + a_3:\r<BR>\n" );
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document.write( "a_1 + a_2 + a_3 = 10 + 10 + 10 = 30<BR>\n" );
document.write( "However, we need to consider the behavior of the function as r approaches 0 or infinity.\r<BR>\n" );
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document.write( "As r approaches 0, 10/r approaches infinity, so S approaches infinity.<BR>\n" );
document.write( "As r approaches infinity, 10r approaches infinity, so S approaches infinity.<BR>\n" );
document.write( "Therefore, the minimum value of a_1 + a_2 + a_3 occurs when r = 1, and the minimum value is 30.\r<BR>\n" );
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document.write( "Final Answer: The smallest possible value of a_1 + a_2 + a_3 is 30. </SPAN>\n" );
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