Question

An AP consist of 37 terms , the sum of the three middle most terms is 429 . find the AP.

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Answer 1

Answer:

middle terms will be a18,a19,a20

a+17d+a+18d+a+19d=429

3a+54d=429

a+18d=143

Step-by-step explanation:

a+17d+143+a+19d=429

2a+36d=286.

I guess there should be some more data .

question is incomplete girl.

Answer 2

Here is the answer to your question.

Here is the answer to your question.Let the first term and the common difference of the A.P are a and d respectively.

Here is the answer to your question.Let the first term and the common difference of the A.P are a and d respectively.Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term= 19th term.

Here is the answer to your question.Let the first term and the common difference of the A.P are a and d respectively.Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term= 19th term.Thus, three middle most terms of this A.P.are 18th, 19th and 20th terms.

Here is the answer to your question.Let the first term and the common difference of the A.P are a and d respectively.Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term= 19th term.Thus, three middle most terms of this A.P.are 18th, 19th and 20th terms.Given a 18 + a 19 + a 20 = 225

Here is the answer to your question.Let the first term and the common difference of the A.P are a and d respectively.Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term= 19th term.Thus, three middle most terms of this A.P.are 18th, 19th and 20th terms.Given a 18 + a 19 + a 20 = 225(a + 17d) + (a + 18d) + (a + 19d) = 225

Here is the answer to your question.Let the first term and the common difference of the A.P are a and d respectively.Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term= 19th term.Thus, three middle most terms of this A.P.are 18th, 19th and 20th terms.Given a 18 + a 19 + a 20 = 225(a + 17d) + (a + 18d) + (a + 19d) = 225→ 3(a + 18d) = 225

Here is the answer to your question.Let the first term and the common difference of the A.P are a and d respectively.Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term= 19th term.Thus, three middle most terms of this A.P.are 18th, 19th and 20th terms.Given a 18 + a 19 + a 20 = 225(a + 17d) + (a + 18d) + (a + 19d) = 225→ 3(a + 18d) = 225⇒ a +18d = 75

Here is the answer to your question.Let the first term and the common difference of the A.P are a and d respectively.Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term= 19th term.Thus, three middle most terms of this A.P.are 18th, 19th and 20th terms.Given a 18 + a 19 + a 20 = 225(a + 17d) + (a + 18d) + (a + 19d) = 225→ 3(a + 18d) = 225⇒ a +18d = 75→ a=75-18d... (1)

Here is the answer to your question.Let the first term and the common difference of the A.P are a and d respectively.Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term= 19th term.Thus, three middle most terms of this A.P.are 18th, 19th and 20th terms.Given a 18 + a 19 + a 20 = 225(a + 17d) + (a + 18d) + (a + 19d) = 225→ 3(a + 18d) = 225⇒ a +18d = 75→ a=75-18d... (1)According to given information

Here is the answer to your question.Let the first term and the common difference of the A.P are a and d respectively.Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term= 19th term.Thus, three middle most terms of this A.P.are 18th, 19th and 20th terms.Given a 18 + a 19 + a 20 = 225(a + 17d) + (a + 18d) + (a + 19d) = 225→ 3(a + 18d) = 225⇒ a +18d = 75→ a=75-18d... (1)According to given informationa 35+ a 36 + a 37 = 429

→ 3(a +35d) = 429

→ 3(a +35d) = 429→ (75-18d) + 35d = 143

→ 3(a +35d) = 429→ (75-18d) + 35d = 14317d=143-75-68

→ 3(a +35d) = 429→ (75-18d) + 35d = 14317d=143-75-68⇒ d = 4

→ 3(a +35d) = 429→ (75-18d) + 35d = 14317d=143-75-68⇒ d = 4Substituting the value of d in equation (1), it is obtained

→ 3(a +35d) = 429→ (75-18d) + 35d = 14317d=143-75-68⇒ d = 4Substituting the value of d in equation (1), it is obtaineda = 75-18 x 4 = 3

→ 3(a +35d) = 429→ (75-18d) + 35d = 14317d=143-75-68⇒ d = 4Substituting the value of d in equation (1), it is obtaineda = 75-18 x 4 = 3Thus, the A.P. is 3, 7, 11, 15...

→ 3(a +35d) = 429→ (75-18d) + 35d = 14317d=143-75-68⇒ d = 4Substituting the value of d in equation (1), it is obtaineda = 75-18 x 4 = 3Thus, the A.P. is 3, 7, 11, 15...Hope! This will help you.

→ 3(a +35d) = 429→ (75-18d) + 35d = 14317d=143-75-68⇒ d = 4Substituting the value of d in equation (1), it is obtaineda = 75-18 x 4 = 3Thus, the A.P. is 3, 7, 11, 15...Hope! This will help you.Cheers!

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