document.write( "Question 1184673: Please help me solve this question
\n" ); document.write( "the fifth, ninth and sixteenth term of a linear sequence(a.p) are consecutive terms of an exponential sequence(g.p)
\n" ); document.write( "(i) find the common difference of the linear sequence in terms of the first terms
\n" ); document.write( "(ii) show that the twenty-first,thirty-seventh and sixty-fifth terms of the linear sequence are consecutive terms of an exponential sequence whose common ratio is 7/4
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Algebra.Com's Answer #815305 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "(i) 5th term of the ap: a+4d
\n" ); document.write( "9th term: a+8d
\n" ); document.write( "16th term: a+15d

\n" ); document.write( "In a gp, the square of any term is equal to the product of the terms before and after it:

\n" ); document.write( "\"%28a%2B8d%29%5E2=%28a%2B4d%29%28a%2B15d%29\"
\n" ); document.write( "\"a%5E2%2B16ad%2B64d%5E2=a%5E2%2B19ad%2B60d%5E2\"
\n" ); document.write( "\"4d%5E2-3ad=0\"
\n" ); document.write( "\"d%284d-3a%29=0\"

\n" ); document.write( "The problem is of little interest if d=0, so

\n" ); document.write( "\"4d-3a=0\"
\n" ); document.write( "\"4d=3a\"
\n" ); document.write( "\"d=%283%2F4%29a\"

\n" ); document.write( "(i) ANSWER: d = (3/4)a

\n" ); document.write( "(ii) 21st term: a+20d = a+15a = 16a
\n" ); document.write( "37th term: a+36d = a+27a = 28a
\n" ); document.write( "65th term: a+64d = a+48a = 49a

\n" ); document.write( "28a/16a = 7/4; 49a/28a = 7/4; this sequence is geometric with common ratio 7/4

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