Question

In an ap Sm/Sn=m⁴/n⁴, then prove that Tm+1/Tn+1=(2m+1)³/(2 +1)³

Answer 1

Answer

Tm+1 / Tn+1 = ( 2m+1 )³ / ( 2n+1 )³

Given

In an AP  $\rm \dfrac{S_m}{S_n}=\dfrac{m^4}{n^4}$

To Prove

$\bullet \;\; \rm \dfrac{T_{m+1}}{T_{n+1}}=\dfrac{(2m+1)^3}{(2n+1)^3}$

Solution

Given that ,

$\to \rm \dfrac{S_m}{S_n}=\dfrac{m^4}{n^4}\\\\\to \rm \dfrac{\frac{m}{2}(2a+(m-1)d)}{\frac{n}{2}(2a+(n-1)d)}=\dfrac{m^4}{n^4}\\\\\to \rm \dfrac{m(2a+(m-1)d)}{n(2a+(n-1)d)}=\dfrac{m^4}{n^4}\\\\\to \rm \dfrac{2a+(m-1)d}{2a+(n-1)d}=\dfrac{m^3}{n^3}$

Let's sub. m=2m+1 and n=2n+1 ,

$\to \rm \dfrac{2a+(2m+1-1)d}{2a+(2n+1-1)d}=\dfrac{(2m+1)^3}{(2n+1)^3}\\\\\to \rm \dfrac{2a+2md}{2a+2nd}=\dfrac{(2m+1)^3}{(2n+1)^3}\\\\\to \rm \dfrac{a+md}{a+nd}=\dfrac{(2m+1)^3}{(2n+1)^3}\\\\\to \rm \dfrac{a_{m+1}}{a_{n+1}}=\dfrac{(2m+1)^3}{(2n+1)^3}\\\\\to \rm \dfrac{T_{m+1}}{T_{n+1}}=\dfrac{(2m+1)^3}{(2n+1)^3}$

Hence proved