# 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. - Mathematics

Sum

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.

#### Solution

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 + 36 d) = 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,….

Concept: Sum of First n Terms of an A.P.
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#### APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 5 Arithematic Progressions
Exercise 5.4 | Q 4 | Page 57
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