document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1153523>Question 1153523</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The first, third and ninth terms of a linear sequence (A.P) are equal to the first three terms of a exponential sequence (G.P). if the seventh term of the linear sequence is 14.<BR>\n" );
document.write( "Calculate:<BR>\n" );
document.write( "(a) the twentieth term of the linear sequence <BR>\n" );
document.write( "(b) the sum of the first twelve term of the G.P.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #775783 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=775783\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Part (a)\r<BR>\n" );
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document.write( "Let {a1,a2,a3,...} be the arithmetic progression (AP) and {b1,b2,b3,...} be the geometric progression (GP)\r<BR>\n" );
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document.write( "If a1 = x, then<BR>\n" );
document.write( "a2 = x+d for some constant d<BR>\n" );
document.write( "a3 = x+2d<BR>\n" );
document.write( "a4 = x+3d<BR>\n" );
document.write( "a5 = x+4d<BR>\n" );
document.write( "a6 = x+5d<BR>\n" );
document.write( "a7 = x+6d = 14, given 7th term of AP<BR>\n" );
document.write( "a8 = x+7d<BR>\n" );
document.write( "a9 = x+8d<BR>\n" );
document.write( "We add on d each time to get the next term.\r<BR>\n" );
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document.write( "x+6d = 14 solves to x = 14-6d<BR>\n" );
document.write( "We'll use this equation later. Let's call this equation (1).\r<BR>\n" );
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document.write( "Since the first, third and ninth terms of the AP are equal to the first three terms of the GP, this means<BR>\n" );
document.write( "a1 = b1<BR>\n" );
document.write( "a3 = b2<BR>\n" );
document.write( "a9 = b3\r<BR>\n" );
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document.write( "The first three terms of the GP are<BR>\n" );
document.write( "b1 = x<BR>\n" );
document.write( "b2 = x*r<BR>\n" );
document.write( "b3 = x*r^2<BR>\n" );
document.write( "We multiply each term by r to get the next term.\r<BR>\n" );
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document.write( "Earlier we found that a3 = x+2d, which equates to b2 to get<BR>\n" );
document.write( "a3 = b2<BR>\n" );
document.write( "x+2d = x*r<BR>\n" );
document.write( "14-6d+2d = (14-6d)*r ..... plug in x = 14-6d, which was equation (1).<BR>\n" );
document.write( "14-4d = (14-6d)*r<BR>\n" );
document.write( "r = (14-4d)/(14-6d) ... divide both sides by (14-6d)<BR>\n" );
document.write( "We'll use this later. Call this equation (2).\r<BR>\n" );
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document.write( "Also,<BR>\n" );
document.write( "a9 = b3<BR>\n" );
document.write( "x+8d = b3 ..... plug in a9 = x+8d<BR>\n" );
document.write( "x+8d = x*r^2 .... plug in b3 = x*r^2<BR>\n" );
document.write( "(14-6d)+8d = (14-6d)*r^2 ... plug in equation (1)<BR>\n" );
document.write( "14+2d = (14-6d)*r^2<BR>\n" );
document.write( "14+2d = (14-6d)*((14-4d)/(14-6d))^2 .... plug in equation (2).<BR>\n" );
document.write( "14+2d = ((14-4d)^2)/(14-6d) .... one pair of (14-6d) terms cancel<BR>\n" );
document.write( "(14+2d)(14-6d) = (14-4d)^2 .... multiply both sides by (14-6d)<BR>\n" );
document.write( "196-84d+28d-12d^2 = 196-56d-56d+16d^2 .... use the FOIL rule to expand<BR>\n" );
document.write( "196-56d-12d^2 = 196-112d+16d^2 ... combine like terms<BR>\n" );
document.write( "0 = 196-112d+16d^2-196+56d+12d^2 .... get everything to one side<BR>\n" );
document.write( "0 = 28d^2-56d ... combine like terms<BR>\n" );
document.write( "28d^2-56d = 0<BR>\n" );
document.write( "28d(d-2) = 0 ... factor<BR>\n" );
document.write( "28d = 0 or d-2 = 0 .... zero product property<BR>\n" );
document.write( "d = 0 or d = 2\r<BR>\n" );
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document.write( "If d = 0, then the AP is {14, 14, 14, 14, 14, 14, ...} which is trivial. Likely your teacher will want you to focus on the more interesting solution.\r<BR>\n" );
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document.write( "If d = 2, then<BR>\n" );
document.write( "x = 14-6d<BR>\n" );
document.write( "x = 14-6*2<BR>\n" );
document.write( "x = 2<BR>\n" );
document.write( "Therefore, a1 = 2 is the first term of the AP.\r<BR>\n" );
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document.write( "The nth term of the AP is<BR>\n" );
document.write( "a(n) = a1 + d(n-1)<BR>\n" );
document.write( "a(n) = 2 + 2(n-1)<BR>\n" );
document.write( "a(n) = 2 + 2n-2<BR>\n" );
document.write( "a(n) = 2n<BR>\n" );
document.write( "A few items of the sequence are: {<font color=blue>2</font>, 4, <font color=blue>6</font>, 8, 10, 12, 14, 16, <font color=blue>18</font>, ...} <BR>\n" );
document.write( "The first, third and ninth terms are highlighted in <font color=blue>blue</font><BR>\n" );
document.write( "Note that the sequence 2,6,18 shows up in part (b) as the first three terms of the GP.\r<BR>\n" );
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document.write( "Plug in n = 20<BR>\n" );
document.write( "a(n) = 2n<BR>\n" );
document.write( "a(20) = 2*20<BR>\n" );
document.write( "a(20) = 40<BR>\n" );
document.write( "The 20th term of the AP is <font color=red>40</font>\r<BR>\n" );
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document.write( "Part (b)\r<BR>\n" );
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document.write( "Use the value of d to calculate r<BR>\n" );
document.write( "r = (14-4d)/(14-6d)<BR>\n" );
document.write( "r = (14-4*2)/(14-6*2)<BR>\n" );
document.write( "r = (14-8)/(14-12)<BR>\n" );
document.write( "r = 6/2<BR>\n" );
document.write( "r = 3\r<BR>\n" );
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document.write( "The first term of the GP is<BR>\n" );
document.write( "b1 = x<BR>\n" );
document.write( "b1 = 2\r<BR>\n" );
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document.write( "Calculate the sum of the first 12 terms of the GP. Starting term is a = 2 and common ratio is r = 3<BR>\n" );
document.write( "Sn = a*(1-r^n)/(1-r)<BR>\n" );
document.write( "S12 = 2*(1-3^12)/(1-3)<BR>\n" );
document.write( "S12 = 2*(1-531441)/(1-3)<BR>\n" );
document.write( "S12 = 2*(-531440)/(-2)<BR>\n" );
document.write( "S12 =  <font color=red>531440</font> \r<BR>\n" );
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document.write( "The long way to get this answer is to generate the first 12 terms<BR>\n" );
document.write( "b1 = 2<BR>\n" );
document.write( "b2 = 6 ... multiply previous term by r = 3 to get the next term<BR>\n" );
document.write( "b3 = 18<BR>\n" );
document.write( "b4 = 54<BR>\n" );
document.write( "b5 = 162<BR>\n" );
document.write( "b6 = 486<BR>\n" );
document.write( "b7 = 1458<BR>\n" );
document.write( "b8 = 4374<BR>\n" );
document.write( "b9 = 13122<BR>\n" );
document.write( "b10 = 39366<BR>\n" );
document.write( "b11 = 118098<BR>\n" );
document.write( "b12 = 354294<BR>\n" );
document.write( "Then add up the terms<BR>\n" );
document.write( "2+6+18+54+162+486+1458+4374+13122+39366+118098+354294 = <font color=red>531440</font> \r<BR>\n" );
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document.write( "</font> </SPAN>\n" );
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