Question

Example 6 The sum of n terms of two arithmetic progressions are in the ratio
(3 + 8): (in + 15). Find the ratio of their 12 terms.​

Answer 1

Correct Question

The sum of n terms of two arithmetic progressions are in the ratio

(3n + 8): (7n + 15). Find the ratio of their 12 terms.​

Solution

Let assume First term (a) , Common Difference(d)

Formula

⇒Tₙ = a+(n-1)d

⇒Sₙ = n/2{2a + (n-1)d}

Now we have to find ratio of 12th term

⇒(T₁₂ of (i)st AP)/(T₁₂ of (ii)st AP)

⇒(a + (12-1)d)/(A + (12-1)D

⇒(a + 11d)/(A + 11D)

Given

The sum of n terms of two arithmetic progressions are in the ratio

⇒(3n + 8):(7n + 15)

⇒(Sₙ of (i)st AP)/(Sₙ of (ii)st AP) = (3n+8)/(7n+15)

⇒ n/2{2a + (n-1)d}/ n/2{2A + (n-1)D} = (3n+8)/(7n+15)

n/2 is cancel , we get

⇒{2a + (n-1)d}/{2A + (n-1)D} = (3n+8)/(7n+15)

⇒2{a + [(n-1)d]/2}/2{A + [(n-1)D]/2} = (3n+8)/(7n+15)

⇒{a + [(n-1)d]/2}/{A + [(n-1)D]/2} = (3n+8)/(7n+15)

Now we find value of n , so we can write as

⇒(n-1)/2=11

⇒(n-1) = 22

⇒n=23

so take n = 23

⇒{a + [(23-1)d]/2}/{A + [(23-1)D]/2} = (3×23+8)/(7×23+15)

⇒(a+11d)/(A+11D) = 77/176

⇒(a+11d)/(A+11D) = 7/16

The ratio of their 12 term id 7/16