Question

If the \[{(m + 1)^{th}},{(n + 1)^{th}}\] and \[{(r + 1)^{th}}\] terms of an A.P. are in G.P. and \[m,n,r\] are in H.P., then the value of the ratio of the common difference to the first term of the A.P. is
A. \[ - \frac{2}{n}\]
B. \[\frac{2}{n}\]
C. \[ - \frac{n}{2}\]
D. \[\frac{n}{2}\]

Answer 1

HINT:
Given that \[{(m + 1)^{th}},{(n + 1)^{th}}\] and \[{(r + 1)^{{\rm{th }}}}\] terms of an A.P. are in G.P. and ‘m’, ‘n’, and ‘r’ are in H.P., we are asked to determine the ratio of the first term and the common difference of the A.P. We are aware that an Arithmetic progression's and a Geometric progression's respective nth terms are \[a + (n - 1)d\] and \[a{r^{n - 1}}\], respectively.
Complete step-by-step solution:
H.P. suggests that the terms' reciprocals are in A.P.
Given that ‘m’, ‘n’, and ‘r’ are in the HP, it follows that \[\frac{1}{m},\frac{1}{n},\frac{1}{r}\] are in the HP.
Terms in A.P. differ in a common way.
Consequently,
\[\frac{1}{n} - \frac{1}{m} = \frac{1}{r} - \frac{1}{n} \Rightarrow \frac{2}{n} = \frac{1}{m} + \frac{1}{r}\]
Assuming that the common difference is d and the first term of the given A.P. is ‘a’,
Then, \[{a_{m + 1}} = a + md,{a_{n + 1}} = a + nd{\rm{\;}}\] and \[{a_{r + 1}} = a + rd\]
 By the given information, \[{(m + 1)^{th}},{(n + 1)^{th}}\]and\[{(r + 1)^{th}}\] terms of an A.P. are in G.P.
We can state that,
\[\frac{{a + nd}}{{a + md}} = \frac{{a + rd}}{{a + nd}} \Rightarrow {(a + nd)^2}\]
\[ = (a + md)(a + rd)\]
Now, there are two expressions here.
They are,
\[2mr = n(r + m){\rm{\;}}\] and \[{(a + nd)^2} = (a + md)(a + rd)\]
Now, solve the above the expressions:
Let’s simplify that,
\[{\left( {1 + n\left( {\frac{d}{a}} \right)} \right)^2} = \left( {1 + m\left( {\frac{d}{a}} \right)} \right)\left( {1 + r\left( {\frac{d}{a}} \right)} \right)\]
Let's assume that \[\frac{d}{a}\], in which case \[1 + nx{)^2} = (1 + mx)(1 + rx)\] will be the result.
Now, we have to expand the above expression, we obtain
\[1 + {n^2}{x^2} + 2nx = 1 + mx + rx + mr{x^2}\]
Write the above expression in terms of factors:
\[ \Rightarrow {x^2}\left( {mr - {n^2}} \right) + x(m + r - 2n) = 0\]
Simplify, we will get two expressions:
\[ \Rightarrow x = 0\]
Or
\[x = \frac{{ - (m + r - 2n)}}{{\left( {mr - {n^2}} \right)}}\]
Here, we have to use the first expression\[2mr = n(r + m)\] in \[\frac{{ - (m + r - 2n)}}{{(mr - {n^2})}}\], we obtain:
\[x = \frac{{ - (m + r - 2n)}}{{\left( {\frac{{n(m + r)}}{2} - {n^2}} \right)}}\]
Simplify the above expression:
\[ \Rightarrow x = \frac{{ - 2}}{n}\]
Therefore, then the value of the ratio of the common difference to the first term of the A.P. is \[x = \frac{{ - 2}}{n}\]
Hence, the option A is correct.
Note:
We should be confident in the concepts we will use in between the steps when responding to questions of this type. If someone hadn't carefully read the question and been instructed that we were asked to determine the ratio of the common difference to the first term, they would have come up with the incorrect response of \[\frac{{ - 2}}{n}\].