Question

Example The sum of n terms of two arithmetic progressions are in the ratio
(3n+8): (7n + 15). Find the ratio of their 12^th terms.​

Answer 1

the sum of n terms of two arithmetic progressions are in the ratio (3n + 8) : (7n + 15) . Find the ratio of their 12th terms.

Answer Text

Let a is the first term and d1 is the common difference of the first AP.

Let b is the first term and d2 is the common difference of the second AP.

Then,

n2(2a+(n−1)d1)n2(2b+(n−1)d2)=3n+87n+15

⇒2a+(n−1)d12b+(n−1)d2=k(3n+8)k(7n+15)

Comparing numerators for both sides,

⇒2a−d1=8k,d1=3k

⇒a=112k

Similarly, comparing denominators of both sides,

⇒2b−d2=15k,d2=7k

⇒b=11k

So, ratio of their 12th terms,

a12b12=a+11d1b+11d2

=112k+33k11k+77k=77k176k=716

So, required ratio is 716.