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प्रश्न
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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उत्तर
Since, total number of terms (n) = 37 ...[odd]
∴ Middle term = `((37 + 1)/2)^("th")` term = 19th term
So, the three middle most terms = 18th, 19th and 20th,
By given condition,
Sum of the three middle most terms = 225
a18 + a19 + a20 = 225
⇒ (a + 17d) + (a + 18d) + (a + 19d) = 225 ...[∵ an = a + (n – 1)d]
⇒ 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 equation (i) from equation (ii), we get
17d = 68
⇒ d = 4
From equation (i),
a + 18(4) = 75
⇒ a = 75 – 72
⇒ 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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