Question

An Arithmetic Progression consists of 3737 terms. The sum of the first 33 terms of it is 1212 and the sum of the last 33 terms is 318318 , then find the first and last terms of the progression ​

Answer 1

Given :-

• The arithmetic progression consists of 37 terms .

Solution :-

• Let us assume that the first term is a and the common difference is d. Then the last term becomes ( a + 36d).

The sum of the first three terms is 12.

Thus,

• a + a + d + a + 2d = 12
• 3a + 3d = 12
• a + d = 4

Now, The sum of the last three terms is 318

• a + 34d + a + 35d + a + 36d = 318
• 3a + 105d = 318
• a + 35d = 106

Subtracting this from the first equation

• 34d = 102
• d = 3
• a = 1

Last term :-

• a + 36d = 1 + 108 = 109

Answer : The first and last terms of the ap are 1 and 109 respectively .

Answer 2

Solution -

It is given that the arithmetic progression consists of 37 terms . Let us assume that the first term is a and the common difference is d. Then the last term becomes ( a + 36d).

The sum of the first three terms is 12.

So

a + a + d + a + 2d = 12

> 3a + 3d = 12

> a + d = 4

Now

The sum of the last three terms is 318

> a + 34d + a + 35d + a + 36d = 318

> 3a + 105d = 318

> a + 35d = 106

Subtracting this from the first equation

> 34d = 102

> d = 3

> a = 1

Last term > a + 36d = 1 + 108 = 109

Answer : The first and last terms of the ap are 1 and 109 respectively .

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