1.

The Sum of the three terms of an A.P. is 21 and the product of the first, and the third terms exceed the second term by 6, find three terms.

Answer»

Given : 

The sum of first three terms is 21 

To find : 

The first three terms of AP 

Assume the first three terms are a - d, a, a + d where a is the first term and d is the common difference 

So, 

Sum of first three terms is a - d + a + a + d = 21 

3a = 21 

a = 7 

It is also given that product of first and third term exceeds the second by 6.

So,

(a - d)(a + d) - a = 6 

a2 - d2 - a = 6

Substituting a = 7,

72 - d2 - 7 = 6 

d2 = 36 

d = 6 

or d = - 6 

Hence,

The terms of AP are a - d, a, a + d which is 1, 7, 13 or 13, 7, 1.



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