Let a1,a2 & d1,d2 be the first terms & common differences of the two arithmetic progressions respectively. According to the given condition, we have

sum‌of‌n‌terms‌of‌first‌A.P.

sum‌of‌n‌terms‌of‌second‌A.P.

=

5n+4

9n+6

⇒

n 2 [2a1+(n−1)d1]

n 2 [2a2+(n−1)d2]

=

5n+4

9n+6

⇒

2a1+(n−1)d1

2a2+(n−1)d2

=

5n+4

9n+6

... (i)
Substituting n = 35 in (i), we get

2a1+34d1

2a2+34d2

=

(5×35)+4

(9×35)+6

⇒

a1+17d1

a2+17d2

=

179

321

∴

18th‌term‌of‌first‌A.P.

18th‌term‌of‌second‌A.P.

=

a1+17d1

a2+17d2

=

179

321