Question

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

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Answer 1

Answer:

If Sn is the sum of the first n term of a sequence, the ratio will be expressed as:

Sn/S'n = 5n+4/9n+6

The sum of nth term of a sequence is expressed as

Sn = n/2[2a+(n-1)d]

The expression becomes

{n/2[2a+(n-1)d]}/{n/2[2a+(n-1)d']} = 5n+4/9n+4

Open the bracket

a+(n-1/2)d/a+(n-1/2)d' = 5n+4/9n+4

Since the 18th tern is expressed as

T18 = a+(18-1)d

T18 = a+17d

The expression becomes

T18/T'18 = a+(n-1/2)d/a+(n-1/2)d'

a+17d/a'+17d' = a+(n-1/2)d/a+(n-1/2)d'

Compare and contrast to get n

On comparing both sides of the equation

17 = n-1/2

n-1 = 17×2

n-1 = 34

n = 34+1

n = 35

If the nth term is 5.n+4

18th term will be 5(35)+4 = 179

If the nth term is 9n+6

18th term will be 9(35)+6= 321

Hence the ratio of their 18th term will be 179:321

Step-by-step explanation:

$Hope \: it \: helps.$

Answer 2

Step-by-step explanation:

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