The ratio of the 11th term to the 18th term of an ap is 2 : 3. Find the ratio of the 5thterm to the 21st term, and also the ratio of the sum of the first five terms to the sumof the first 21 terms.
Answers
Answer:
1) 1 / 3
2) 5 / 98
Step-by-step explanation:
11th term = a + 10d
18th term = a + 17d
(a + 10d) / (a + 17d) = 2 / 3
3(a + 10d) = 2(a + 17d)
3a + 30d = 2a + 34d
a = 4d -------------> (1)
5th term = a + 4d
Substituting 1,
5th term = 4d + 4d = 8d
21st term = a + 20d
Substituting 1,
21st term = 4d + 20d = 24d
5th term : 21st term = 8d / 24d
= 8 / 24
= 1 / 3
Sum of first 5 terms = 5 / 2 x (2a + 4d)
Substituting 1,
S(5) = 5 / 2 x (8d + 4d)
= 5 / 2 x 12d
= 30d
Sum of first 21 terms = 21 / 2 x (2a + 20d)
Substituting 1,
S(21) = 21 / 2 x (8d + 20d)
= 21 / 2 x (28d)
= 294d
S(5) : S(21) = 30d / 294d
= 5 / 98
Hope this helps! :)
If it does please mark it as brainliest!
Answer:
1) 1 / 3
2) 5 / 98
Step-by-step explanation:
11th term = a + 10d
18th term = a + 17d
(a + 10d) / (a + 17d) = 2 / 3
3(a + 10d) = 2(a + 17d)
3a + 30d = 2a + 34d
a = 4d -------------> (1)
5th term = a + 4d
Substituting 1,
5th term = 4d + 4d = 8d
21st term = a + 20d
Substituting 1,
21st term = 4d + 20d = 24d
5th term : 21st term = 8d / 24d
= 8 / 24
= 1 / 3
Sum of first 5 terms = 5 / 2 x (2a + 4d)
Substituting 1,
S(5) = 5 / 2 x (8d + 4d)
= 5 / 2 x 12d
= 30d
Sum of first 21 terms = 21 / 2 x (2a + 20d)
Substituting 1,
S(21) = 21 / 2 x (8d + 20d)
= 21 / 2 x (28d)
= 294d
S(5) : S(21) = 30d / 294d
= 5 / 98
Hope this helps! :)
If it does please mark it as brainliest!