Arithmetic and Geometric ProgressionsDate: 03/23/98 at 11:24:38 From: ELIZABETH ANGOL Subject: Arithmetic progression If the third term of an arithmetic progression is three times the seventh term and the ninth term is 1, find the first term, the common differences, and the first term less than 0. The third term of a geometric progression is 36 and the fifth term is 16. Find the tenth term. Date: 03/23/98 at 14:22:47 From: Doctor Rob Subject: Re: Arithmetic progression An arithmetic progression has the form a, a+d, a+2*d, a+3*d, a+4*d, ..., where a is the first term, d is the common difference, and the n-th term is a + (n-1)*d. The conditions of your problem translate into: a+2*d = 3*(a+6*d) a+8*d = 1 You need to solve these two simultaneous linear equations for a and d. Once you have done that, you need to substitute them into the inequality a+(n-1)*d < 0, and solve for n. If I have done this problem right, the simultaneous equations are inconsistent, that is, there is no common solution (a, d), so the second part of the question is meaningless. --------------------------------------------------------------------- A geometric progression has the form a, a*r, a*r^2, a*r^3, a*r^4, ..., where a is the first term, r is the common ratio, and the n-th term is a*r^(n-1). The conditions of your problem translate into: a*r^2 = 36 a*r^4 = 16 and you are asked to find a*r^9. One way to do that is to solve these equations for a and r. A good method is to divide the first equation into the second, and discover that r^2 = 4/9, giving two values of r. Now divide that equation into the first, giving you the value of a. Now you can compute a*r^9. There are two answers. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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