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Question
The sum of 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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Solution
In the given problem, the sum of three terms of an A.P is 21 and the product of the first and the third term exceeds the second term by 6.
We need to find the three terms
Here,
Let the three terms be (a - d), a, (a + d) where, a is the first term and d is the common difference of the A.P
So,
`(a - d) + a + (a + d) = 21`
3a = 21
a = 7 .....(1)
Also,
(a - d)(a + d) = a + 6
`a^2 - d^2 = a + 6`
`a^2 - d^2 = a + 6 ("Using " a^2 - b^2 = (a + b)(a - b))`
`(7)^2 - d^2 = 7 + 6` (Using 1)
`49 - 13 = d^2`
Further solving for d
`d^2 = 36`
`d = sqrt36`
d = 6 .....(2)
Now using the values of a and d the expression of the three terms,we get
First term = a - d
So
a - d = 7 - 6
= 1
Second term = a
So
a + d = 7 + 6
= 13
Therefore the three tearm are 1, 7 and 13
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