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Question 4
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.

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Solution

Total number of terms (n) = 37
Middle term=(37+12)th term=19thterm
So, the three middle most terms are 18th, 19th and 20th terms.
By the given condition,
Sum of the three middle most terms = 225
a18+a19+a20=225
(a+17d)+(a+18d)+(a+19d)=225
3a+54d=225
a+18d=75 ...(i)
And,
Sum of the last three terms = 429

a35+a36+a37=429
(a+34d)+(a+35d)+(a+36d)=429
3a+105d=429
a+35d=143 ...(ii)
On subtracting Eq. (i) from Eq.(ii), we get,
17d = 68
d = 4
From Eq.(i),
a + 18(4) = 75

a=7572
a=3
Required AP is a, a + d, a + 2d, a + 3d
i.e., 3, 3+4, 3+2(4), 3+3(4)
i.e., 3, 7, 3+8, 3+12….
i.e., 3, 7, 11, 15….

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