The sum of `n`terms of two arithmetic progressions are in the ratio `(3n+8):(7n+15)dot`Find the ratio of their 12th terms.

The sum of `n`terms of two arithmetic progressions are in the ratio `(

1.	The sum of `n`terms of two arithmetic progressions are in the ratio `(3n+8):(7n+15)dot`Find the ratio of their 12th terms.
Answer» Let ` a_(1),a_(2) and d_(1) , d_(2)` be the first terms and common differences of the first and second AP respectively. Then, by the given condition. we have ` (" sum to n terms of first AP")/( " sum to n terms of second AP")= ( 3n +8)/( 7n +15) ` ` Rightarrow (n/2 .[2a_(1) + ( n-1) d_(1)])/ (n/2. [2a_(2) + (n-1) d_(2)]) = ( 3n +8)/( 7n +15)` ` Rightarrow ( 2a_(1) + (n -1) d_(1))/( 2a_(2) + (n-1) d_(2) )= ( 3n+8)/( 7n + 15) ` Now, ( " 12 th term of first AP")/( " 12th term of second AP") = ( a_(1) +11d_(1))/ (a_(2) +11d_(1))` ` (2a_(1) +22d_(1))/ (2a_(2) +22d_(2))= ( 2a_(1) + (23-1)d_(1))/(2a_(2) + (23-1)d_(2))` ` (3xx 23+8)/ 7 xx 23 +15) ` [ putting n=23in (i)] ` 77/ 176 = 7/16` Hence, the ratio of the 12th terms of given APs is 7:16