The sum of n tearm of two arithmetic progressions are in the ratio (7n+1):(4n+27).Find the ratio of their 11th terms
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To find the ratio of the 11th terms between the two arithmetic progressions, we need to determine the sum of the first n terms for each progression.
Let's start by finding the sum of the first n terms for the first arithmetic progression. The formula for the sum of the first n terms of an arithmetic progression is given by:
Sn = (n/2)(2a + (n-1)d)
Where:
- Sn is the sum of the first n terms
- a is the first term of the arithmetic progression
- d is the common difference between the terms
Let's assume the first arithmetic progression has a first term of a1 and a common difference of d1. Therefore, the sum of the first n terms for this progression is:
S1 = (n/2)(2a1 + (n-1)d1)
Similarly, for the second arithmetic progression with a first term of a2 and a common difference of d2, the sum of the first n terms is:
S2 = (n/2)(2a2 + (n-1)d2)
Given that the ratio of the sum of the first n terms is (7n+1):(4n+27), we can express this as an equation:
S1/S2 = (7n+1)/(4n+27)
Now, let's substitute the expressions for S1 and S2:
[(n/2)(2a1 + (n-1)d1)] / [(n/2)(2a2 + (n-1)d2)] = (7n+1)/(4n+27)
Canceling out the common terms, we get:
(2a1 + (n-1)d1) / (2a2 + (n-1)d2) = (7n+1)/(4n+27)
Now, we need to find the ratio of their 11th terms. let's denote the 11th term of the first progression as T1 and the 11th term of the second progression as T2.
The 11th term of an arithmetic progression is given by:
Tn = a + (n-1)d
Therefore, we have:
T1 = a1 + (11-1)d1
T2 = a2 + (11-1)d2
We need to find the ratio T1/T2.
Finally, let's substitute the expressions for T1 and T2 into the equation:
(T1/T2) = (a1 + 10d1)/(a2 + 10d2)
This provides the ratio of the 11th terms of the two arithmetic progressions.
for 1st series:
first term is a, common difference is d
sum(n) = (n/2)(2a + d(n-1))
for 2nd series:
first term is b, common difference is e
sum(n) = (n/2)(2b + e(n-1))
(n/2)(2a + d(n-1)) / (n/2)(2b + e(n-1)) = (7n+1)/(4n+27)
(2a + dn - d)/ (2b + en- e) = (7n+1)/(4n+27)
8an + 54a + 4dn^2 + 27dn - 4nd - 27d = 14bn + 7en^2 + 2b + en - e ***
You also know that n ≥ 11, and n must be a whole number
ratio of 11 terms = (a+10d)/(b+10e) **
Does that give you something to play with ?