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in Arithmetic Progression by (48.7k points)
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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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We know that,

First term of an AP = a

Common difference of AP = d

nth term of an AP, an = a + (n – 1)d

Since, n = 37 (odd),

Middle term will be (n+1)/2 = 19th term

Thus, the three middle most terms will be,

18th, 19th and 20th terms

According to the question,

a18 + a19 + a20 = 225

Using an = a + (n – 1)d

a + 17d + a + 18d + a + 19d = 225

3a + 54d = 225

3a = 225 – 54d

a = 75 – 18d … (1)

Now, we know that last three terms will be 35th, 36th and 37th terms.

According to the question,

a35 + a36 + a37 = 429

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

3a + 105d = 429

a + 35d = 143

Substituting a = 75 – 18d from equation 1,

75 – 18d + 35d = 143 [ using eqn1]

17d = 68

d = 4

Then,

a = 75 – 18(4)

a = 3

Therefore, the AP is a, a + d, a + 2d….

i.e. 3, 7, 11….

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