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Home » 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.
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.
Arithmetic Progression | CBSE | Class 10 | Exercise 5.4 | Maths | NCERT Exemplar
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:

We all know that,

First term of an AP = a

The common difference = d

AP’s nth term of an, {{a}_{n}}~=a+\left( n1 \right)d

Since, n\text{ }=\text{ }37 (odd),

Middle term will be \left( n+1 \right)/2\text{ }=\text{ }{{19}^{th~}}term

As a result, the three middle most terms will be, 18th, 19th and 20th terms

According to the question,

{{a}_{18}}~+\text{ }{{a}_{19}}~+\text{ }{{a}_{20}}~=\text{ }225

Using {{a}_{n}}~=\text{ }a+\left( n1 \right)d

a+17d+a+18d+\text{ }a+19d=225

3a+54d=225

3a\text{ }=\text{ }22554d

a\text{ }=\text{ }7518d… (1)

We now know that the 35th, 36th, and 37th terms will be the last three.

According to the question,

{{a}_{35}}~+\text{ }{{a}_{36}}~+\text{ }{{a}_{37}}~=\text{ }429

a\text{ }+\text{ }34d\text{ }+\text{ }a\text{ }+\text{ }35d\text{ }+\text{ }a\text{ }+\text{ }36d\text{ }=\text{ }429

3a\text{ }+\text{ }105d\text{ }=\text{ }429

a\text{ }+\text{ }35d\text{ }=\text{ }143

Substituting a\text{ }=\text{ }7518d from equation 1,

75\text{ }\text{ }18d\text{ }+\text{ }35d\text{ }=\text{ }143 [ using eq.1]

17d\text{ }=\text{ }68

d\text{ }=\text{ }4

Then,

a\text{ }=\text{ }7518\left( 4 \right)

a\text{ }=\text{ }3

As a result, the AP is a, a + d, a + 2d….

i.e. 3, 7, 11….

i

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