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Question: In a geometric sequence, t7 = -192 and t15 = - 49152. Find "a" and "r".

In a geometric sequence, each term is formed by multiplying the previous term with a constant, denoted by r (the common ratio). The form for the nth term, tn, in a geometric sequence is given by tn = ar^n-1, where a is the first term, r is the common ratio, and n is the position of the term. Given t7 = -192 and t15 = -49152, we can form two equations: ar^6 = -192 -----------(Equation 1) ar^14 = -49152 -----------(Equation 2) To express r in terms of a, we could divide Equation 2 by Equation 1, remembering that division in exponential terms translates to subtraction of the powers: r^14 / r^6 = -49152/-192 r^8 = 256 Taking the cube-root of both sides, we find r = 2 (since r could be negative, we take only the positive root considering the general trend of the sequence). We substitute r = 2 into the Equation 1: a * (2)^6 = -192, a * 64 = -192, a = -192 / 64 = -3. Therefore, a = -3 and r = 2.

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