An AP consist of 37 terms, the sum of the three middle most term is 225 and the sum of the last three terms of AP is 429...
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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 it's correct and helps to you....
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