# An AP consists of 37 terms. The sum of the three middle most terms is 225 and thesum 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) + (+ 18d) + (+ 19d) = 225
⇒ 3(a + 18d) = 225
⇒ a + 18d = 75
⇒ a = 75 – 18d … (1)

According to given information
a 35 + a 36 + a 37 = 429
⇒ (+ 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 …

Cheers!

• 351

the three middle terms would be the 18th, 19th and 20th terms

let these terms be a+ 17d + a + 18d + a + 19d = 225

=> 3a + 54d = 225

=> a + 18d = 75.......i

Sum of the last 3 terms = 429

a + 36d + a+ 35d + a+ 34d = 429

3a + 105d = 429

a + 35d = 143.........ii

subtracting ii from i

17d= 68

d = 4

substituting in i

a= 3

hence the ap is 3, 7,11..............

• 105
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