Question #247051

Find, showing all working, a formula for the n-th term tn of the sequence (tn) defined by

t1 = 5; tn = -7tn-1 /3, n >= 2.


1
Expert's answer
2021-10-06T14:07:09-0400
t1=5,tn=7tn13t_1=5, t_n=-\dfrac{7t_{n-1}}{3}

tntn1=7tn13tn1=73\dfrac{t_n}{t_{n-1}}=\dfrac{-\dfrac{7t_{n-1}}{3}}{t_{n-1}}=-\dfrac{7}{3}

We have a geometric progression (tn)(t_n)


t1=5,q=73t_1=5, q=-\dfrac{7}{3}

Then


tn=5(73)n1,n1t_n=5(-\dfrac{7}{3})^{n-1}, n\geq1


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