Question

If n

a.m are inserted between 20 and 80 such that the rayio of first mean to the last mean is 1:3 then find the value of n

Answer 1

★ ARITHMETIC MEAN ★

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Let $A_{1}, A_{2}, A_{3}, .....A_{n}$ be n arithmetic means between 20 and 80
And let d be the common difference between the terms of the A.P.

So, $d = \frac{b - a}{n + 1}$
          $= \frac{80 - 20}{n + 1}$
          $= \frac{60}{n + 1}$

Now, A₁ $= 20 + d$

              $= 20 + \frac{60}{n + 1}$
              $= 20(\frac{n + 4}{n + 1})$

And, $A_{n} = 20 + d$
                  $= 20 + \frac{60n}{n + 1}$
                  $= 20(\frac{4n + 1}{n + 1})$

Thus, $\frac{ A_{1} }{ A_{n} } = \frac{1}{3}$

⇒ $\frac{20(\frac{n + 4}{n + 1})}{20 (\frac{4n + 1}{n + 1})} = \frac{1}{3}$

⇒ $\frac{n + 4}{4n + 1} = \frac{1}{3}$
⇒ $4n + 1 = 3n + 12$
⇒ $n = 11$

∴ The value of n is 11. 

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#ExoticExplorer

Answer 2

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