Question

The ratio between the sum of `n^(th)` term is of two arithmetic progressions is `(7n+1)/(4n+27)`. Find the ratio of their eleventh term?

Answer 1

`a_11/A_11=4/3`

Explanation:

Let us take two arithmetic sequence :

`(i)a,a+d,a+2d,a+3d,...,a+(n-1)d,...`

`(ii)A,A+D,A+2D,A+3D,...,A+(n-1)D,...`

So, sum of `n^(th)term :`

`S_n=n/2[2a+(n-1)d] and S'_n=n/2[2A+(n-1)D]`

`:.ratio =(n/2[2a+(n-1)d])/(n/2[2A+(n-1)D])=(7n+1)/(4n+27)`

`=>(2a+(n-1)d)/(2A+(n-1)D)=(7n+1)/(4n+27)...to(1)`

Now,

ratio of `n^(th)term=a_n/A_n=(a+(n-1)d)/(A+(n-1)D`

ratio of `11^(th)term=a_11/A_11=(a+(11-1)d)/(A+(11-1)D`

`i.e.a_11/A_11=(a+10d)/(A+10D)`

`=>a_11/A_11=(2a+20d)/(2A+20D)...to(2)`

Comparing `LHS,of (1) and RHS, of (2)` we can say that

`n-1=20=>n=21`

So, we take `n=21` into `(1)`

`(2a+(21-1)d)/(2A+(21-1)D)=(7(21)+1)/(4(21)+27).`

`=>(2a+20d)/(2A+20D)=(147+1)/(84+27)`

`=>a_11/A_11=148/111...to`from`(2)`

`=>a_11/A_11=4/3`