document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=195215>Question 195215</A>: <I>A Special rubber ball is dropped from a height of 128 meters . Each time the ball bounces to half its previous height. After 1 bounce the ball reaches a height of 64 meters.\r<BR>\n" );
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document.write( "a. How high did the ball reach after the 7th bounce.<BR>\n" );
document.write( "b.How high does the ball get after the 9th bounce</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #146449 by </B><A HREF=/tutors/aboutme.mpl?userid=windsolace><B>windsolace(8)</B></A><IMG SRC=http://www.algebra.com/images/stars/stars-05.gif>&nbsp;<A HREF=/tutors/aboutme.mpl?userid=windsolace><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=146449\">put this solution on YOUR website!</A><BR>The series identified is a geometric progression with common ratio of half.<BR>\n" );
document.write( "Formula for geometric progression:\r<BR>\n" );
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document.write( "Nth term of series = ar^(n-1), where a = first term value, r = common ratio\r<BR>\n" );
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document.write( "a) 7th term of the series = 64(1/2)^(7-1) = 64(1/2)^6<BR>\n" );
document.write( "                                           = 64(1/64)<BR>\n" );
document.write( "                                           = 1\r<BR>\n" );
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document.write( "The ball reached a height of 1m after the 7th bounce\r<BR>\n" );
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document.write( "b) 9th term of series = 64(1/2)^8<BR>\n" );
document.write( "                      = 64/256<BR>\n" );
document.write( "                      = 0.250\r<BR>\n" );
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document.write( "The ball reached a height of 0.25m (25cm) after the 9th bounce.</TD></TR></TABLE>\n" );