The sum of n terms of two arithmetic series
are in the ratio of 7n+1/4n+27 find the ratio of their 12th terms
Answers
Step-by-step explanation:
Given :-
The sum of n terms of two arithmetic series
are in the ratio of 7n+1/4n+27
To find :-
Find the ratio of their 12th terms ?
Solution :-
We Know that
In an AP , The sum of first n terms is denoted by Sn and defined by (n/2)[2a+(n-1)d]
Let the first term and Common difference of the first AP be a and d
Let the first term and the common difference of the second AP be b and c
Sum of n terms of the first AP
=> Sn = (n/2)[2a+(n-1)d] ------(1)
Sum of first n terms of the second AP
=>Sn = (n/2)[2b+(n-1)c] --------(2)
Their ratio
=> (n/2)[2a+(n-1)d] : (n/2)[2b+(n-1)c]
=> (n/2)[2a+(n-1)d] / (n/2)[2b+(n-1)c]
=> [2a+(n-1)d] / [ 2b+(n-1)c]
According to the given problem
The sum of n terms of two arithmetic series
are in the ratio = 7n+1/4n+27
=> [2a+(n-1)d] / [ 2b+(n-1)c] = (7n+1)/(4n+27)
To find ratio of 12th terms we can replace n by 23 because (2×12-1) = 24-1 = 23
=> [2a+(23-1)d]/[2b+(23-1)c] = [7(23)+1]/[4(23)+27]
=>(2a+22d)/(2b+22c) = (161+1)/(92+27)
=> (2a+22d)/(2b+22c) = 162/119
=> 2(a+11d)/2(b+11c) = 162/119
=>(a+11d)/(b+11c) = 162/119
=> a12 /b12 = 162/119
=> a 12 : b 12 = 162:119
Therefore ratio of 12 th terms = 162:119
Answer:-
The ratio of 12th terms of both AP's is 162:119
Used formulae:-
In an AP , The sum of first n terms is denoted by Sn and defined by (n/2)[2a(n-1)d]
nth term = an = a+(n-1)d
- a = First term
- d = Common difference
- n = Number of terms