Question

An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP.

Answer 1

The middlemost terms are the 18th, 19th, and the 20th terms.
Hence, a18 + a19 +a20 = 225
Or, a+17d + a+ 18d + a + 19d = 225
Or, 3a + 54d =225 .......(i)

Now, a35+a36+a37 = 429
Or, a+34d+a+35d+a+36d = 429
Or, 3a+105d = 429 ........(ii)

Eleminating 3a from i and ii, we get,
105d-54d = 429-225
Or, 51d = 201
Or, d = 4 (approx)
Or, a = 3

Hence, the AP is 3,7,11,15,.........

Answer 2

Answer:

An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.

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

a 35 + a 36 + a 37 = 429

⇒ (a + 34d) + (a + 35d) + (a + 36d) = 429

⇒ 3(a + 35d) = 429

⇒ (75 – 18d) + 35d = 143

⇒ 17d = 143 – 75 = 68

⇒ d = 4

Substituting the value of d in equation (1), it is obtained

a = 75 – 18 × 4 = 3

Thus, the A.P. is 3, 7, 11, 15 …

Hope! This will help you.

Cheers!