Question

n arithmetic means are inserted between 3 and 17. If the ratio of the last and the first arithmetic mean is 3:1, then the value of n is:
(a) 9
(b) 6
(c) 7
(d) 5

Answer 1

Hint: First of all insert n arithmetic numbers between 3 and 17 and get the arithmetic series as \[3,{{A}_{1}},{{A}_{2}},{{A}_{3}}.....17\]. Now, this series contains (n + 2) terms. So, use the formula for the nth term of the A.P that is \[{{a}_{n}}=a+\left( n-1 \right)d\] to write (n+2)the term that is 17. Now, use \[\dfrac{{{A}_{n}}}{{{A}_{1}}}=\dfrac{3}{1}\] to get another equation in n and d. Finally solving both the equations to get the value of n.

Complete step-by-step answer:
We are given that n arithmetic means are inserted between 3 and 17. Also, if the ratio of the last and first arithmetic mean is 3: 1, we have to find the value of n. First of all, let us consider that there are n arithmetic means between 3 and 17 and they are \[{{A}_{1}},{{A}_{2}},{{A}_{3}},{{A}_{4}}....{{A}_{n}}\]. So, we get the arithmetic series as:

\[3,{{A}_{1}},{{A}_{2}},{{A}_{3}},{{A}_{4}}.....{{A}_{n}},17\]

Now, we can see that the above series has total (n + 2) terms and let the common difference

of the above series be d. We know that the nth term of A.P is

\[{{a}_{n}}=a+\left( n-1 \right)d.....\left( 1 \right)\]

We know that, in the first term of AP = 3, (n + 2)th term of A.P = 17, the common difference of A.P = d.

By substituting the values of a = 3, d = d, n = (n + 2) in equation (1), we get,

\[{{a}_{n+2}}=3+\left( n+2-1 \right)d\]

\[\Rightarrow {{a}_{n+2}}=3+\left( n+1 \right)d\]

We know that \[{{a}_{n+2}}=17\]. So, we get,

\[\Rightarrow 17=3+\left( n+1 \right)d\]

\[\Rightarrow \left( n+1 \right)d=17-3=14\]

So, we get,

\[\left( n+1 \right)d=14....\left( i \right)\]

Also, we are given that the ratios of the last and the first arithmetic mean is 3:1. So, we get,

\[\dfrac{{{A}_{n}}}{{{A}_{1}}}=\dfrac{3}{1}.....\left( ii \right)\]

We know that \[{{A}_{1}}\] is the 2nd term of the given series and \[{{A}_{n}}\] is (n + 1)th

term of the series. So, by using equation (1), we get,

\[{{A}_{n}}=a+\left( n+1-1 \right)d\]

\[=3+\left( n \right)d\]

\[{{A}_{1}}=a+\left( 2-1 \right)d\]

\[=3+d\]

By substituting the values of \[{{A}_{n}}\] and \[{{A}_{1}}\] in equation (ii), we get,

\[\Rightarrow \dfrac{3+nd}{3+d}=\dfrac{3}{1}\]

By cross multiplying the above equation, we get,

\[\left( 3+nd \right)1=3\left( 3+d \right)\]

\[\Rightarrow 3+nd=9+3d\]

\[\Rightarrow nd=9-3+3d=6+3d\]

So, we get,

\[nd=6+3d\]

\[\Rightarrow n=\dfrac{6+3d}{d}.....\left( iii \right)\]

By substituting the value of n in equation (i), we get,

\[\left( \dfrac{6+3d}{d}+1 \right)d=14\]

By simplifying the above equation and canceling the like terms, we get,

\[\left( 6+3d+d \right)=14\]

\[\Rightarrow 6+4d=14\]

\[\Rightarrow 4d=14-6\]

\[\Rightarrow 4d=8\]

So, we get d = 2.

By substituting the values of d = 2 in equation (iii), we get,

\[n=\dfrac{6+3\left( 2 \right)}{2}\]

\[n=\dfrac{6+6}{2}=\dfrac{12}{2}\]

So, we get n = 6.

Hence, option (b) is the right answer.

Note: Students must remember the general nth term of A.P that is \[{{a}_{n}}=a+\left( n-1 \right)d\] and note that this is valid for all values of n. Also, here apart from writing \[{{A}_{n}}=3+nd\], students could take \[{{A}_{n}}=17-d\] and can directly solve for d by substituting it in \[\dfrac{{{A}_{n}}}{{{A}_{1}}}=\dfrac{3}{1}\] without the use of the other equation. Some students make this mistake of considering \[{{A}_{1}}\] as the first term and \[{{A}_{n}}\] as the nth term which is wrong. If we see the series clearly, we can say that \[{{A}_{1}}\] is the 2nd term and \[{{A}_{n}}\] is (n + 1)th term of A.P.