Question

If the sums of n terms of two arithmetic progressions are in the ratio \[\frac{3n + 5}{5n - 7}\] , then their n^th terms are in the ratio

  

Options

• \[\frac{3n - 1}{5n - 1}\]

   

• \[\frac{3n + 1}{5n + 1}\]

   

• \[\frac{5n + 1}{3n + 1}\]

   

• \[\frac{5n - 1}{3n - 1}\]

   

Answer 1

In the given problem, the ratio of the sum of n terms of two A.P’s is given by the expression,

`(S_n)/(S'_n) = (3n + 5)/(5n+ 7)`                  ....(1)

We need to find the ratio of their n^th terms.

Here we use the following formula for the sum of n terms of an A.P.,

`S_n = n/2 [ 2a + ( n - 1)d]`

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So,

`S_n = n/2 [ 2a + ( n - 1)d]`

Where, a and d are the first term and the common difference of the first A.P.

Similarly,

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

Where, a^’ and d^’ are the first term and the common difference of the first A.P.

So,

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

       `= ([2a + (n-1)d])/([2a'+(n-1)d'])`              ............(2)

Equating (1) and (2), we get,

`= ( [2a + (n-1)d])/( [2a'+(n-1)d'])=(3n + 5)/(5n + 7)`

Now, to find the ratio of the n^th term, we replace n by2n - 1 . We get,

`= ( [2a + (n-1-1)d])/( [2a'+(n-1-1 )d'])= (3(2n - 1) + 5)/(5(2n -1) + 7)`

` ( 2a + (2n-2)d)/( 2a'+(2n-2)d')= (6n - 3 + 5 )/(10n - 5 + 7)`

` ( 2a + 2(n-1)d)/( 2a'+2(n-1)d')=(6n +2)/(10n + 2)`

` ( a + (n-1)d)/( a'+(n-1)d') = (3n + 1)/(5n + 1)`

As we know,

a_n = a + (n - 1)d 

Therefore, we get,

`a_n /(a'_n) = (3n + 1) /(5n +1)`