The sum of first nth terms of two APs are in the ratio (3n+8) is to (7n+15).Find the ratio of their 12th terms

Sum of first n terms of two AP's are in the ratio (3n + 8) : (7n + 15).

the ratio of their 12th terms

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.