Question

There are 37 terms in an AP the sum of three terms placed exactly at middle is 225 and sum of last three terms is 429. write an AP

Answer 1

Hey

Here is your answer,

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

So,

(a – d) + a + (a + d) = 225
3a = 225
a = 75

Now, the AP is
a – 18d, …, a – 2d, a – d, a, a + d, 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
d = 4

Now, first term = a – 18d = 75 – 18(4) = 3
Therefore, The AP is 3, 7, 11, … , 147.Here

Hope it helps you!

Answer 2

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 …