Question

the sum of three consecutive terms of an ap is 3 and the product is negative 35 find the numbers

Answer 1

Let the numbers be a - d, a, a + d
Sum will be
a - d + a + a + d = 3
3a = 3
a = 1

(a - d) (a) (a + d) = - 35
(a square - d square) (1) = - 35
1x1 - d square = - 35
1 - d square = - 35
- d square = - 36
d square = 36
d = 6
The numbers are - 5, 1, 7

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Answer 2

Let the AP be (a-d),(a),(a+d) where (a-d) is the first term and d is the common difference
Given
$a + (a + d )+ (a - d) = 3 \\ = > 3a = 3 \\ = > a = \frac{3}{3} = 1$
So The Middle Term is 1
$a(a - d)(a + d) = - 35 \\ as \: a = 1 \: \: so \\ 1(1 - d)(1 + d) = - 35 \\ = > {1}^{2} - {d}^{2} = - 35 \\ = > 35 + 1 = {d}^{2} \\ = > d = \sqrt{36} \\ = > d = 6$
So the terms are
(a-d), (a) and (a+d)
=(1-6) , (1) and (1+6)
= -5 , 1 , 7
$so \: the \: numbers \: are \:$
-5, 1 and 7

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