Question

There are n arithmetic mean between 3 and 17 . The ratio of the first mean and the last is 1:3 .Find n

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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

Let A_{1},A_{2},A_{3},.........A_{n} be the n arithmetic means between 3 and 17.

Then, the common difference is given by:

d=\frac{b-a}{n+1}

=\frac{17-3}{n+1}

d=\frac{14}{n+1}

Now, A_{1}=a+d=3+\frac{14}{n+1}

=\frac{3n+3+14}{n+1}=\frac{3n+17}{n+1}

and A_{n}=a+nd=3+\frac{14n}{n+1}

=\frac{17n+3}{n+1}

Now, \frac{A_{n}}{A_{1}}=\frac{3}{1}

\frac{17n+3}{3n+17}=\frac{3}{1}

17n+3=9n+51

n=6