Question

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.
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Answer 1

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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

a^2 - d^2 - a = 6

substituting a = 7

7^2 - d^2 - 7 = 6

d^2 = 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

Answer 2

Answer:-

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

a^2 - d^2 - a = 6

substituting a = 7

7^2 - d^2 - 7 = 6

d^2 = 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