Question

There are ‘n’ A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3:1 .find the value of n.

Answer 1

Hint: We have given n arithmetic means between 3 and 17 that means we have an AP of n+2 terms where first term is 3 and last term is 17. And use the data given in question to proceed further.

Complete step-by-step answer:
Let ‘n’ A.M.s be ${A_1},{A_2},{A_3}, \ldots ,{A_n}$
So we have the given A.P is
$3,{A_1},{A_2}, \ldots ,{A_n},17$
In our case we have
${a_n} = 17,a = 3$ and number of terms = (n+2)
$\because {a_n} = a + \left( {n - 1} \right)d$
$17 = 3 + \left[ {\left( {n + 2} \right) - 1} \right]d \\\ 17 - 3 = \left( {n + 1} \right)d \\\ \therefore d = \dfrac{{14}}{{\left( {n + 1} \right)}} \\\$
Now
${A_1} = a + d = 3 + \dfrac{{14}}{{n + 1}} \\\ {A_n} = a + nd = 3 + n\dfrac{{14}}{{n + 1}} \\\$
As given in question,
$\dfrac{{{A_n}}}{{{A_1}}} = \dfrac{3}{1} \\\ \Rightarrow \dfrac{{3 + \dfrac{{14n}}{{n + 1}}}}{{3 + \dfrac{{14}}{{n + 1}}}} = 3 \\\$
\Rightarrow \dfrac{{3\left( {n + 1} \right) + 14n}}{{3\left( {n + 1} \right) + 14}} = 3 \\\ \Rightarrow \dfrac{{17n + 3}}{{3n + 17}} = 3 \\\ \Rightarrow 17n + 3 = 3\left( {3n + 17} \right). \\\ \Rightarrow 17n + 3 = 9n + 51 \\\ \Rightarrow 8n = 48 \Rightarrow n = 6 \\\

Note: Whenever you get this type of question the key concept of solving is you have to solve like an AP of n+2 terms and simple mathematics to use the data given in question and use the formula of finding the last term of AP to get an answer.