Question

In an A.P. if $ \dfrac{{{S_m}}}{{{S_n}}} = \dfrac{{{m^4}}}{{{n^4}}} $ then prove that $ \dfrac{{{T_{m + 1}}}}{{{T_{n + 1}}}} = \dfrac{{{{(2m + 1)}^3}}}{{{{(2n + 1)}^3}}} $

Answer 1

Hint: This question is based on arithmetic progression and this is also a sequence of the form $ a,a + d,a + 2d,... $ where a is the first term and d is the common difference. Here in this question the relation of the sum of the arithmetic progress is given and we have to prove that it is inter linked to the nth term of the progression. To prove the above inequality, we need the sum of A.P formula and the nth term of A.P.

Complete step-by-step answer:
The general arithmetic progression is of the form $ a,a + d,a + 2d,... $ where a is first term nth d is the common difference. The nth term of the arithmetic progression is defined as $ {T_n} = a + (n - 1)d $ and the sum of the arithmetic progression is defined as $ {S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] $
By the given data we have $ \dfrac{{{S_m}}}{{{S_n}}} = \dfrac{{{m^4}}}{{{n^4}}} $ ………………… (1)
As we know that $ {S_m} = \dfrac{m}{2}\left[ {2a + (m - 1)d} \right] $ and $ {S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] $
Substituting these inequalities in equation (1) we have
 $ \dfrac{{{S_m}}}{{{S_n}}} = \dfrac{{{m^4}}}{{{n^4}}} $
 $ \Rightarrow \dfrac{{\dfrac{m}{2}\left[ {2a + (m - 1)d} \right]}}{{\dfrac{n}{2}\left[ {2a + (n - 1)d} \right]}} = \dfrac{{{m^4}}}{{{n^4}}} $
On simplification we have
 $ \Rightarrow \dfrac{{m\left[ {2a + (m - 1)d} \right]}}{{n\left[ {2a + (n - 1)d} \right]}} = \dfrac{{{m^4}}}{{{n^4}}} $
Cancel the similar terms on both sides we have
 $ \Rightarrow \dfrac{{\left[ {2a + (m - 1)d} \right]}}{{\left[ {2a + (n - 1)d} \right]}} = \dfrac{{{m^3}}}{{{n^3}}} $
In the LHS divide both numerator and denominator by 2 we have
 $ \Rightarrow \dfrac{{\left[ {a + \dfrac{{(m - 1)}}{2}d} \right]}}{{\left[ {a + \dfrac{{(n - 1)}}{2}d} \right]}} = \dfrac{{{m^3}}}{{{n^3}}} $
Let we substitute as $ \dfrac{{m - 1}}{2} = M $ and $ \dfrac{{n - 1}}{2} = N $ ………………. (2)
After substitution we have
 $ \Rightarrow \dfrac{{\left[ {a + Md} \right]}}{{\left[ {a + Nd} \right]}} = \dfrac{{{m^3}}}{{{n^3}}} $
The numerator and the denominator of LHS is of the form (M+1) th term and (N+1) th term
So we write the above inequality as
 $ \Rightarrow \dfrac{{{T_{M + 1}}}}{{{T_{N + 1}}}} = \dfrac{{{m^3}}}{{{n^3}}} $ …………………………. (3)
Now let us consider the equation (2) we have alter the terms we have
 $ \dfrac{{m - 1}}{2} = M \Rightarrow m - 1 = 2M \Rightarrow m = 2M + 1 $ and $ \dfrac{{n - 1}}{2} = N \Rightarrow n - 1 = 2N \Rightarrow n = 2N + 1 $
On substitution of above inequalities to the equation (3) we have
 $ \Rightarrow \dfrac{{{T_{M + 1}}}}{{{T_{N + 1}}}} = \dfrac{{{{(2M + 1)}^3}}}{{{{(2N + 1)}^3}}} $
We will replace M as m and N as n then we have
 $ \dfrac{{{T_{m + 1}}}}{{{T_{n + 1}}}} = \dfrac{{{{(2m + 1)}^3}}}{{{{(2n + 1)}^3}}} $
Hence we proved.
So, the correct answer is “ $ \dfrac{{{T_{m + 1}}}}{{{T_{n + 1}}}} = \dfrac{{{{(2m + 1)}^3}}}{{{{(2n + 1)}^3}}} $ ”.

Note: We must know about the arithmetic progression arrangement and it is based on the first term and common difference. The common difference of the arithmetic progression is defined as $ {a_2} - {a_1} $ where $ {a_2} $ represents the second term and $ {a_1} $ represents the first term. The sum of n terms and the nth term formula is necessary to prove the above inequality.