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 a₁, a₂ and d₁, d₂ 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

2a₁ + (n - 1) d₁/2a₂ + (n - 1) d₂

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

Substituting n = 35 in (1) , we obtain

⇒ 2a₁ + (34) d₁/2a₂ + (34) d₂

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

⇒ a₁ + 17d₁/a₂ + 17d₂

= 179/321 ....(2)

Now,

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

= a₁+ 17d₁/a₂+ 17d₂

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

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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