The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18^th terms.

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Solution

Let a₁, a₂, and d₁, d₂be the first terms and the common difference of the first and second arithmetic progression respectively.

According to the given condition,

Substituting n = 35 in (1), we obtain

From (2) and (3), we obtain

Thus, the ratio of 18^th term of both the A.P.s is 179: 321.

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5 n + 4 : 9 n + 6

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The sums of ' $n$ ' terms of two arithmetic progression are in the ration $5 n + 4 : 9 n + 6$ , find the ration of their $18^{t h}$ terms.

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