Terms are there. Divide 51 into three parts in AP so that the largest exceeds the smallest by 10.​

The first three terms of the Arithmetic Progression are 12, 17 and 22.

For an ARITHMETIC PROGRESSION, AP ;

First term = a

Second term = a + d

Third term = a + 2d

Where, d = common difference ;

If third term exceeds smallest by 10 ;

Third term - first term

a + 2d - a = 10

2d = 10

d = 10/2

d = 5

Sum of the three terms :

a + (a + d) + a + 2d = 51

3a + 3d = 51

d = 5

3a + 3(5) = 51

3a + 15 = 51

3a = 51 - 15

3a = 36

a = 12

The AP would be:

First term, a = 12

Second term, a + d = 12 + 5 = 17

Third term = a + 2(d) = 12 + 10 = 22

Therefore , the first three terms of the AP are :

12, 17 and 22

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