Given:
Sum of first n terms of two AP's are in the ratio (3n + 8) : (7n + 15).
Find:
the ratio of their 12th terms
Solution:
Let the first terms and common differences of the two APs be a , d & a' , d'.
We know that,
Sum of first n terms of an AP (Sn) = n/2 * [ 2a + (n - 1)d ]
So,
⟹ n/2 * [ 2a + (n - 1)d ] : n/2 * [ 2a' + (n - 1)d' ] = 3n + 8 : 7n + 15
⟹ 2a + (n - 1)d : 2a' + (n - 1)d' = 3n + 8 : 7n + 15
Multiply numerator and denominator by 1/2 in LHS
⟹ a + (n - 1)/2 * d : a' + (n - 1)/2 * d' = 3n + 8 : 7n + 15
Now,
3n + 8 : 7n + 15 is the ratio of nth terms of the two APs.
We have to find the ratio of 12th terms.
We know that,
nth term of an AP (aₙ) = a + (n - 1)d
⟹ a₁₂ = a + (12 - 1)d
⟹ a₁₂ = a + 11d
So,
⟹ a + (n - 1)/2 * d = a + 11d
⟹ n - 1/2 = 11
⟹ n - 1 = 22
⟹ n = 23
Hence, we have to substitute n as 23 to find the ratio of their 12th terms.
⟹ a + 11d : a' + 11d' = 3(23) + 8 : 7(23) + 15
⟹ a + 11d : a' + 11d' = 77 : 176
⟹ a + 11d : a' + 11d' = 7 : 16
∴ The ratio of the 12th terms of the two APs is 7 : 16.
I hope it will help you.
Regards.