The sums of n terms of two arithmetic series are in ratio of 2n+1:2n-1. Find the ratio of their 10th terms.
Let n=19. Then we have
a1+a19 = a1+a1+18d = 2(a1+9d) = 2a10
b1+b19 = 2b10
Now, using the sum ratio,
(a1+a19)/(b1+b19) = a10/b10 = (2*19+1)/(2*19-1) = 39/37
To solve this problem, let's start by assuming the first arithmetic series has a common difference of 'a' and the second arithmetic series has a common difference of 'd'.
The sum of the first n terms of an arithmetic series is given by Sn = (n/2)(2a + (n - 1)d).
Given that the sums of the n terms are in the ratio of (2n + 1):(2n - 1), we can set up the equation:
[(n/2)(2a + (n - 1)d)] / [(n/2)(2a + (n - 1)d)] = (2n + 1)/(2n - 1)
Next, we cross multiply to eliminate the fractions:
(2n + 1)(n/2)(2a + (n - 1)d) = (2n - 1)(n/2)(2a + (n - 1)d)
Simplifying both sides, we can cancel out (n/2) terms:
(2n + 1)(2a + (n - 1)d) = (2n - 1)(2a + (n - 1)d)
Expanding both sides:
4an + 2a(n - 1) + (n - 1)(2a + (n - 1)d) = 4an - 2a + (n - 1)(2a + (n - 1)d)
Simplifying further, we can eliminate some terms:
2a + nd - d + nd - d^2 = -2a
Combining like terms:
4a - 2d + 2nd - d^2 = 0
Rearranging the equation:
2nd - d^2 = -4a + 2d
Since this equation holds for all n, the coefficients of each term must be equal. Therefore, we can equate them:
2n = -4
d = 2a
Solving the first equation for n:
2n = -4
n = -2
Since a term index cannot be negative, we discard this solution.
With the second equation, we have d = 2a.
The ratio of the 10th terms can be found using the formula for the nth term in an arithmetic series:
Tn = a + (n - 1)d
For the first arithmetic series:
T1 = a + (1 - 1)d = a
T10 = a + (10 - 1)d = a + 9d = a + 9(2a) = a + 18a = 19a
For the second arithmetic series:
T1 = a + (1 - 1)d = a
T10 = a + (10 - 1)d = a + 9d = a + 9(2a) = a + 18a = 19a
Therefore, the ratio of their 10th terms is 19a:19a, which simplifies to 1:1.
To find the ratio of the 10th terms of two arithmetic series, we need to find the individual terms of each series and then calculate their ratio.
Let's first assume that the first arithmetic series has a common difference of 'a' and the second arithmetic series has a common difference of 'b'. We can then express the sum of the first 'n' terms of these series as follows:
Sum of the first 'n' terms of the first series = n/2 [2a + (n-1)a]
Sum of the first 'n' terms of the second series = n/2 [2b + (n-1)b]
The given information states that the ratios of these sums are (2n+1):(2n-1). Therefore, we have the equation:
(n/2 [2a + (n-1)a]) / (n/2 [2b + (n-1)b]) = (2n+1) / (2n-1)
To simplify this equation, we can cancel out the common terms and simplify the expression:
(2a + (n-1)a) / (2b + (n-1)b) = (2n+1) / (2n-1)
Simplifying further:
(2a +na - a) / (2b + nb - b) = (2n+1) / (2n-1)
(na + a) / (nb + b) = (2n+1) / (2n-1)
Cross-multiplying:
(2n+1)(nb + b) = (2n-1)(na + a)
Expanding both sides:
2n^2b + 2nb + nb + b = 2n^2a - 2na + na + a
Combining like terms:
2n^2b + 3nb + b = 2n^2a - na + a
Now, we need to equate the coefficients of like powers of 'n' on both sides of the equation:
2b = -a
3b = -a
b/a = -1/2
This gives us the ratio of the common differences: b/a = -1/2
Finally, we can find the ratio of the 10th terms by using the common differences:
Ratio of 10th terms = (a + 9a) / (b + 9b) = 10a / 10b = a / b
Substituting the value of b/a = -1/2:
Ratio of 10th terms = a / b = a / (-1/2) = -2a
Therefore, the ratio of the 10th terms of the two arithmetic series is -2a.