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

**Solution**:

Let a_{1}, a_{2} and d_{1}, d_{2} be the first terms and common differences of two arithmetic progressions respectively.

According to the given condition,

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

= 5n + 4/9n + 6

n/2 [2a1 + (n - 1) d1]/n/2 [2a2 + (n - 1) d2]

= 5n + 4/9n + 6

2a_{1} + (n - 1) d_{1}/2a_{2} + (n - 1) d_{2}

= 5n + 4/9n + 6 ....(1)

Substituting n = 35 in (1) , we obtain

⇒ 2a_{1} + (34) d_{1}/2a_{2} + (34) d_{2}

= [5(35) + 4]/[9(35) + 6]

⇒ a_{1} + 17d_{1}/a_{2} + 17d_{2}

= 179/321 ....(2)

Now,

18^{th }term of first A.P/18^{th }term of second A.P

= a_{1}+ 17d_{1}/a_{2}+ 17d_{2}

From (2) and (3), we obtain

18^{th }term of first A.P/18^{th }term of second A.P

= 179/321

Thus, the ratio of their 18th terms is 179: 321

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Question 9

## 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

**Summary:**

Given that the sums of n terms of two arithmetic progression are in the ratio 5n + 4: 9n+ 6, the ratio of their 18th terms in 179:321

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