 # 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.

Given

n = 37

the sum of the three middlemost terms is 225

the sum of the last three is 429

Find out

We have to determine the AP

Solution

We know that the nth term of an AP is given by the formula

an = a + (n – 1)d

where,

a = first term

an is the nth term

d is the common difference

According to the given details

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 given data

a18 + a19 + a20 = 225

We know that an = a + (n – 1)d

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

3a + 54d = 225

3a = 225 – 54d

a = 75 – 18d … (1)

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

a = 75 – 18(4)

a = 3

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

i.e. 3, 7, 11….

The AP obtained is 3, 7, 11…. (0) (0)