CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If the ratio of sum of m terms and n terms of an A.P. be $$m^2 : n^2$$, then the ratio of its $$m^{th}$$ and $$n^{th}$$ terms will be


A
2m-1 : 2n-1
loader
B
m:n
loader
C
2m+1 : 2n+1
loader
D
None
loader

Solution

The correct option is D 2m-1 : 2n-1
By using $$S_{n}=\dfrac{n}{2}\left [ 2a+(n-1)d \right ]$$ we have

Given that $$\dfrac { \dfrac { m }{ 2 } [2a+(m-1)d] }{ \dfrac { n }{ 2 } [2a+(n-1)d] } =\dfrac { { m }^{ 2 } }{ { n }^{ 2 } } $$

$$\Rightarrow \dfrac { 2a+(m-1)d }{ 2a+(n-1)d } =\dfrac { { m } }{ { n } } $$
$$\Rightarrow \dfrac { a+\dfrac { 1 }{ 2 } (m-1)d }{ a+\dfrac { 1 }{ 2 } (n-1)d } =\dfrac { { m } }{ { n } } $$

$$\Rightarrow an+\dfrac { 1 }{ 2 } (m-1)dn=am+\dfrac { 1 }{ 2 } (n-1)md$$

$$\Rightarrow a(n-m)+\dfrac { d }{ 2 } [mn-n-mn+m]=0$$

$$\Rightarrow a(n-m)+\dfrac { d }{ 2 } (m-n)=0$$

$$\Rightarrow a=\dfrac { d }{ 2 } $$ or $$d=2a$$

Therefore, $$\dfrac { { T }_{ m } }{ { T }_{ n } } =\dfrac { a+(m-1)d }{ a+(n-1)d } $$

$$=\dfrac { a+(m-1)2a }{ a+(n-1)2a }$$

$$=\dfrac {1+2m-2 }{ 1+2n-2}$$ 

$$=\dfrac { 2m-1 }{ 2n-1 } $$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image