Question

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.

Answer 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

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!

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let the three middle most terms of the AP be a - d, a, a + d.

We have , (a - d) + a + (a + d) = 225

⇒ 3a = 225

⇒ a = 225/3

⇒ a = 75

Now, the AP is,

a - 18d,.....,a - 2d,a - d,a,a + 2d,......a + 18d

Sum of last three terms

⇒ (a + 18d) + (a + 17d) + (a + 16d) = 429

⇒ 3a + 51d = 429

⇒ a + 17d = 143

⇒ 75 + 17d = 143

⇒ 17d = 143 - 75

⇒ 17d = 68

⇒ d = 68/17

⇒ d = 4

Now, First term = a - 18d = 75 - 18(4) = 3

Hence, The AP is 3, 7, 11, ..., 147.