The sums of n terms of two arithmetic progressions are in the ratio 5n+4: 9n+6. Find the ratio of their 18th terms.

As an experienced tutor registered on UrbanPro, I can assure you that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into solving the problem at hand.

Given that the sums of nn terms of two arithmetic progressions are in the ratio 5n+4:9n+65n+4:9n+6, we need to find the ratio of their 18th terms.

For an arithmetic progression, the sum of the first nn terms is given by Sn=n2[2a+(n−1)d]Sn=2n[2a+(n−1)d], where aa is the first term and dd is the common difference.

So, for the first arithmetic progression, let's denote its first term as a1a1 and common difference as d1d1, and for the second arithmetic progression, let's denote its first term as a2a2 and common difference as d2d2.

Given that the sum of nn terms of the first arithmetic progression is 5n+45n+4, we have: n2[2a1+(n−1)d1]=5n+42n[2a1+(n−1)d1]=5n+4 Similarly, for the second arithmetic progression with sum 9n+69n+6, we have: n2[2a2+(n−1)d2]=9n+62n[2a2+(n−1)d2]=9n+6

We are given that these sums are in the ratio 5n+4:9n+65n+4:9n+6. Therefore, we can write: 5n+49n+6=n2[2a1+(n−1)d1]n2[2a2+(n−1)d2]9n+65n+4=2n[2a2+(n−1)d2]2n[2a1+(n−1)d1]

Now, simplifying this equation and solving for a2a1a1a2 will give us the ratio of their 18th terms. But first, let's solve for a1a1 and a2a2 by eliminating nn from the given ratios.

5n+49n+6=5n+49n+69n+65n+4=9n+65n+4 n2[2a1+(n−1)d1]n2[2a2+(n−1)d2]=5n+49n+62n[2a2+(n−1)d2]2n[2a1+(n−1)d1]=9n+65n+4 2a1+(n−1)d12a2+(n−1)d2=5n+49n+62a2+(n−1)d22a1+(n−1)d1=9n+65n+4

Now, let's simplify this expression. We'll replace nn with 18, as we are interested in the ratio of their 18th terms. Then, we'll solve for a2a1a1a2 to find the ratio.