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The sum of the n terms of two arithmetic progression are in the ratio (3n+8):(7n+15). Find the ratio of their 12th terms
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Our APs (arithmetic progressions) are
a, a+d, a+2*d, a+3*d, . . .
b, b+e, b+2*e, b+3*e, . . .
where "a" and "b" are first terms and "d" and "e" are their common differences.
In order for to answer the question, it is enough to know three ratios , and .
So, our goal now is to find these three ratios.
1) At n= 1 we have
= = = . (1)
2) At n= 2 we have
= = , or
29*(a+d) = 14*(b+e),
29a + 29d - 14e = 14b.
Divide both sides by "b"
29*(a/b) + 29*(d/b) - 14*(e/b) = 14.
Substitute here = from (1) and multiply both sides by 2. You will get then
58*(d/b) - 28*(e/b) = -1. (2)
3) At n= 3 we have
= = , or
36*(a+2d) = 17*(b+2e),
36a + 72d - 34e = 17b.
Divide both sides by "b"
36*(a/b) + 72*(d/b) - 34*(e/b) = 17.
Substitute here = from (1). You will get then
72*(d/b) - 34*(e/b) = -1. (3)
4) Introduce new variables D = and E = . For these unknowns, from (2) and (3) you have this system of equations
58*D - 28*E = -1 (2')
72*D - 34*E = -1 (3')
Solve it by any method you want / (you know). The solution is D = , E = . Thus = , = .
5) Now we are in position to calculate = = = = = . ANSWER
Solved.