document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1193867>Question 1193867</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Correction: An infinite geometric series has the first term u1 = a and u2 = (1/4a^2)-3a, where a > 0.\r<BR>\n" );
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document.write( "a. Find the values of a for which the sum to infinity of the series exists.<BR>\n" );
document.write( "b. Find the value of a when s infinity = 76. </SPAN>\n" );
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document.write( "Part A\r<BR>\n" );
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document.write( "u1 = first term = a<BR>\n" );
document.write( "u2 = second term = (1/4)a^2 - 3a\r<BR>\n" );
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document.write( "r = common ratio <BR>\n" );
document.write( "r = (u2)/(u1)<BR>\n" );
document.write( "r = ((1/4)a^2 - 3a)/(a)<BR>\n" );
document.write( "r = (a((1/4)a - 3))/(a)<BR>\n" );
document.write( "r = (1/4)a - 3<BR>\n" );
document.write( "where 'a' is nonzero, to avoid a division by zero error.\r<BR>\n" );
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document.write( "For the infinite geometric sum to converge to a finite value, we need -1 < r < 1 to be true.<BR>\n" );
document.write( "The common ratio needs to be between -1 and 1, excluding both endpoints.\r<BR>\n" );
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document.write( "-1 < r < 1<BR>\n" );
document.write( "-1 < (1/4)a - 3 < 1<BR>\n" );
document.write( "-1+3 < (1/4)a - 3+3 < 1+3 ... see note 1 below<BR>\n" );
document.write( "2 < (1/4)a < 4<BR>\n" );
document.write( "2*4 < 4*(1/4)a < 4*4 ... see note 2<BR>\n" );
document.write( "8 < a < 16\r<BR>\n" );
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document.write( "note1: I added 3 to all sides to undo the -3 in the middle<BR>\n" );
document.write( "note2: I multiplied all sides by 4 to undo the 1/4<BR>\n" );
document.write( "note3: The inequality signs do not flip for any of the steps\r<BR>\n" );
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document.write( "Answer: <font color=red>8 < a < 16</font><BR>\n" );
document.write( "The value 'a' needs to be between 8 and 16, excluding both endpoints, so that we converge to a finite sum.\r<BR>\n" );
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document.write( "========================================================<BR>\n" );
document.write( "Part B\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large S_{\infty}\"> is the notation talking about the <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large S_{n}\"> partial sum where n approaches infinity.<BR>\n" );
document.write( "So informally we can say<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large \displaystyle \lim_{n\to\infty}S_n = S_{\infty}\">\r<BR>\n" );
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document.write( "I'll simply refer to it as S to make things a bit more simple.<BR>\n" );
document.write( "This infinite sum S is only possible if 8 < a < 16 found back in part A earlier.\r<BR>\n" );
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document.write( "If that inequality holds up, then it leads to -1 < r < 1 and it allows us to use this infinite geometric sum formula<BR>\n" );
document.write( "S = a/(1-r)\r<BR>\n" );
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document.write( "Let's plug in S = 76 and solve for 'a'<BR>\n" );
document.write( "S = a/(1-r)<BR>\n" );
document.write( "76 = a/(1-r)<BR>\n" );
document.write( "76(1-r) = a<BR>\n" );
document.write( "a = 76(1-r)<BR>\n" );
document.write( "a = 76-76r\r<BR>\n" );
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document.write( "Now plug in r = (1/4)a - 3 and solve for 'a'<BR>\n" );
document.write( "a = 76-76r<BR>\n" );
document.write( "a = 76-76( (1/4)a-3 )<BR>\n" );
document.write( "a = 76-19a + 228<BR>\n" );
document.write( "a+19a = 76 + 228<BR>\n" );
document.write( "20a = 304<BR>\n" );
document.write( "a = 304/20<BR>\n" );
document.write( "a = 76/5<BR>\n" );
document.write( "a = 15.2\r<BR>\n" );
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document.write( "This value of 'a' satisfies the inequality 8 < a < 16, so this is a valid value to allow the infinite geometric series to converge to a finite sum.\r<BR>\n" );
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document.write( "Answer: <font color=red>a = 76/5 = 15.2</font><BR>\n" );
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