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Arithmetic Progression?
AP Question.
The 18th term is twice its 9th term. Find the ratio of the sum of the first 18 terms to the sum of the first 9 terms.
Thanks
Ahh, genius, thanks divide.
2 Answers
- gôhpihánLv 71 decade agoFavorite Answer
formula:
the nth term of an arithmetic progression is:
T (n) = a + (n - 1)d
where a is the first term
d is the common difference
the sum of nth term of an arithmetic progression is:
S (n) = (n/2) (2a + (n - 1)d)
***************
T(18) = 2 T(9)
a + 17d = 2(a + 8d)
a + 17d = 2a + 16d
-a = -d
a = d
S (18)
= (18/2) (2a + 17d)
= 9(2a + 17d)
= 9(2a + 17a)
= 9(19a)
= 171a
S (9)
= (9/2)(2a + 8d)
= 9(a + 4d)
= 9(a + 4a)
= 9(5a)
= 45a
ratio of the sum of the first 18 terms to the sum of the first 9 terms is:
S (18) ÷ S (9)
= 171a ÷ 45a
= 171 ÷ 45
= 19/5
- cidyahLv 71 decade ago
Let a be the first term and d be the common difference.
18th term = a+17d
9th term = a+8d
a+17d = 2(a+8d)
a+17d = 2a+16d
a-d=0
a=d
sum of first 18 terms = 18/2 (2a+17d)
sum of first 9 terms = 9/2 (2a+8d)
dividing 2(2a+17d) / (2a+8d) =2(2a+17d) / (2a+8d)
substitute a=d
2(19d) / 10d = 38/10 = 3.8
