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

Let a₁ and d₁ be the first term and the common difference of the first arithmetic

progression respectively and a₂ and d₂ be the first term and the common difference of the second arithmetic progression respectively.

According to the conditions given in the question,

Sum of n terms of first A.P / Sum of n terms of second A.P.

\( \dfrac{5n+4}{9n+6}\)

 \(\dfrac{\dfrac{n}{2}[2a_1+(n-1)d_1]}{\dfrac{n}{2}[2a_2+(n-1)d_2]}= \dfrac{5n+4}{9n+6}\)

\(\dfrac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}= \dfrac{5n+4}{9n+6} \)

Substitute n = 35  in the equation.

\( \dfrac{2a_1+34d_1}{2a_2+34d_2}= \dfrac{5(35)+4}{9(35)+6}\)

= \( \dfrac{a_1+17d_1}{a_2+17d_2}= \dfrac{179}{321}\)

We also know that 18^th term of first A.P. /18^th term of second A.P.

\( \dfrac{a_1+17d_1}{a_2+17d_2}\)

Then,

18^th term of first A.P. /18^th term of second A.P.= \( \dfrac{179}{321}\)

Therefore, 179 : 321 is the ratio of 8^th term of both the arithmetic progressions.