Question

The sum of n terms of two A.P.'s are in the ratio 5n + 9 : 9n + 6. Then, the ratio of their 18th term is

विकल्प

• \[\frac{179}{321}\]

   

• \[\frac{178}{321}\]

   

• \[\frac{175}{321}\]

   

• \[\frac{176}{321}\]

   

• non above these

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) = (5n + 9 ) /(9n + 6) `                .............(1)  

We need to find the ratio of their 18^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']) = (5n + 9 ) /(9n + 6) ` 

Now, to find the ratio of the n^th term, we replace n by  2n - 1 . We get,

 `=(  [2a + ( 2n-1 - 1) d ])/( [ 2a' + ( 2n- 1- 1) d']) = (5(2n - 1) + 9)/(9(2n - )+6)` 

                `(  2a + (2n - 2) d )/( 2a' + ( 2n- 2) d')= (10n -5 + 9)/(18n-9+6)` 

              `(  2a + 2 (n - 1) d )/(  2a' + 2 ( n- 1) d') = (10n + 4)/(18n - 3)`  

                   `(a + ( n - 1)  d) / (a' + (n - 1) d') = ( 10n + 4)/(18n - 3)`

As we know,

a_n = a + ( n - 1 ) d 

Therefore, for the 18^th terms, we get,

`a_18 /(a'_(18))= (10(18) + 4) / (18(18)- 3) `

            `= 184/321`

Hence  `a_18/(a'_(18)) =184/321`