Question

The sum of n terms of two arithmetic progressions is in the ratio 5n+4:9n+6. Find the ratio of their 18th terms.
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Answer 1

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Let the 2 series have the sum S(1,n) and S(2,n).
now, S(1,n):S(2,n) = (5n+4):(9n+6)
= {5(n-1)+5+4}:{9(n-1)+9+6}
= {2*(9/2)+5(n-1)}:{2*(15/2)+9(n-1)}

= [(n/2){2*(9/2)+5(n-1)}]:[(n/2){2*(15/2)+9(n-1)}] 
hence nth terms will be in the ratio
T(1,n):T(2,n)= [(9/2)+5(n-1)]:[(15/2)+9(n-1)]

=>T(1,13):T(2,13)=[(9/2)+5(13-1)]:[(15/2)+9(13-1)]
=(129/2):(231/2)
=43/77

Answer 2

We know that sum of n terms of an AP sn = (n/2)[2a + (n - 1) * d]

Given sum of n terms of two AP is in ratio 5n + 4: 9n + 6.

Let a,a' be the first terms of the two given AP and d,d' be their respective differences.

⇒ (sum of n terms of 1st AP)/(sum of n terms of 2nd AP) = (5n + 4)/(9n + 6)

$= > \frac{\frac{n}{2}[2a + (n - 1)d]}{\frac{n}{2}[2a' + (n - 1)d']} = \frac{5n + 4}{9n + 6}$

$= > \frac{2a + (n - 1)d}{2a' + (n - 1)d'} = \frac{5n + 4}{9n + 6}$

$= > \frac{a + \frac{n - 1}{2}d}{a' + \frac{n - 1}{2} d'} = \frac{5n + 4}{9n + 6}$

Now,

Given : Ratio of 18th term = (a + 17d)/(a' + 17d').

Substitute (n - 1)/2  = 17

= > (n - 1) = 34

= > n = 35.

Now, substitute n = 35 in above equation.

$= > \frac{a + \frac{35 - 1}{2}d}{a' + \frac{35 - 1}{2} d'} = \frac{5(35) + 4}{9(35) + 6}$

$= > \frac{a + 17d}{a' - 17d'} =\frac{175 + 4}{315 + 6}$

$= > \frac{a + 17d}{a' + 17d'} =\frac{179}{321}$

Therefore, Required ratio = 179:321.

Hope this helps!