0Arithmetic Progression Example
Q: The sums of n terms of two arithmetic progressions are in the ratio 5n+4:9n+6. Find the ratio of their 18th terms.
Solution.
Let a1,a2 and d1,d2 be the first terms and common differences of the first and the second arithmetic progression respectively. Now, as per the problem:
Sum of the n terms of the first AP 5n + 4
-------------------------------------------------- = -------------------
Sum of the n terms of the second AP 9n+16
or,
n/2[2a1+ (n-1)d1] 5n+4
----------------------------------- = ---------------------------------
n/2[2a2+ (n-1)d2] 9n+16
or,
[2a1+ (n-1)d1] 5n+4
----------------------------------- = ------------------------------ ……..(1)
[2a2+ (n-1)d2] 9n+16
Now,
18th Term of first AP a1+17d1
----------------------------------- = -------------------------------- ………(2)
18th Term of second AP a2+17d2
Multiplying Numerator and Denominator of (2) by 2, we have
2a1+34d1 5 * 35 + 4
-------------------------- = ----------------
2a2+34d2 9 * 35 + 6
Therefore, we have,
a2+17d2 179
-------------------------- = ----------------------------------------------
a2+17d2 321
Thus, the required ratio is 179 / 321.
