in a geometric series t1=23,t3=92 and the sum of all of the terms of the series is 62813. How many terms are in the series

Answers

Answer 1

Answer: $t_1=23,\ \ \ t_3=92\n\nthe\ geometric\ series\ \ \ \Rightarrow\ \ \ t_3=t_1\cdot q^2;\ \ \ q- the\ quotient \n\n92=23\cdot q^2\ /:23 \ \ \Rightarrow\ \ \ q^2= 4\ \ \ \Leftrightarrow\ \ \ (q=2\ \ \ or\ \ \ q=-2)\n\nSum\nS_n=t_1\cdot \frac{\big{1-q^n}}{\big{1-q}} \ \ \ \Rightarrow\ \ \ 62813=23\cdot \frac{\big{1-q^n}}{\big{1-q}} \ /:23 \ \ \ \Rightarrow\ \ \2731= \frac{\big{1-q^n}}{\big{1-q}} \n\n$

$1)\ \ \ q=2\n\Rightarrow\ \ \ 2731= \frac{\big{1-2^n}}{\big{1-2}} \ \ \ \Rightarrow\ \ \ 2731=2^n-1\ \ \ \ \Rightarrow\ \ \ 2^n=2732\n\n\Rightarrow\ \ \ n=log_22732\ \notin\ Natural\n\n$

$2)\ \ q=-2\ \ \ \ \Rightarrow\ \ \ 2731= \frac{\big{1-(-2)^n}}{\big{1-(-2)}} \ \ \ \Rightarrow\ \ 2731=\frac{\big{1-(-2)^n}}{\big{3}}\ /\cdot3 \n\n\ \ \ \Rightarrow\ \ \ 8193=1-(-2)^n\ \ \ \Rightarrow\ \ \ (-2)^n=-8192\ \ \ (\Leftrightarrow\ \ \ n-odd\ number)\n\n\ \Rightarrow \ \ \ (-2)^n=(-2)^(13)\ \ \ \Leftrightarrow\ \ \ n=13\n\nAns.\ In\ the\ geometric\ series\ are\ 13\ terms.$

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Answers

Answer:

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Step-by-step explanation:

we know that

1 mi=1,609.34 m ------> 1 m=1/ 1,609.34 mi

1 h= 3,600 sec ----> 1 sec=1/3,600 h

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Convert to mi/h

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Answer:

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Answers

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