The sum of three numbers in A.P. is \[3\] and their product is \[-35\]. Find the numbers.

The sum of three numbers in A.P. is

$3$

and their product is

$-35$

. Find the numbers.

The sum of three numbers in A.P. is

$3$

and their product is

$-35$

. Find the numbers.

From the question it is given that,

The sum of three numbers in A.P. =

$3$

Given, Their product =

$-35$

Let us assume the

$3$

numbers which are in A.P. are, a – d, a, a + d

Now adding

$3$

numbers = a – d + a + a + d =

$3$

$\begin{array}{*{35}{l}} 3a\text{ }=\text{ }3 \\ a\text{ }=\text{ }3/3 \\ a\text{ }=\text{ }1 \\ \end{array}$

From the question, product of

$3$

numbers is

$-35$

$\begin{array}{*{35}{l}} So,\text{ }\left( a\text{ }\text{ }d \right)\text{ }\times \text{ }\left( a \right)\text{ }\times \text{ }\left( a\text{ }+\text{ }d \right)\text{ }=\text{ }\text{ }35 \\ \left( 1\text{ }\text{ }d \right)\text{ }\times \text{ }\left( 1 \right)\text{ }\times \text{ }\left( 1\text{ }+\text{ }d \right)\text{ }=\text{ }\text{ }35 \\ {{1}^{2}}~\text{ }{{d}^{2}}~=\text{ }\text{ }35 \\ {{d}^{2}}~=\text{ }35\text{ }+\text{ }1 \\ {{d}^{2}}~=\text{ }36 \\ d\text{ }=\text{ }\surd 36 \\ d\text{ }=\text{ }\pm 6 \\ \end{array}$

Therefore, the numbers are (a – d) =

$1\text{ }\text{ }6\text{ }=\text{ }\text{ }5$

$\begin{array}{*{35}{l}} a\text{ }=\text{ }1 \\ \left( a\text{ }+\text{ }d \right)\text{ }=\text{ }1\text{ }+\text{ }6\text{ }=\text{ }7 \\ If\text{ }d\text{ }=\text{ }\text{ }6 \\ \end{array}$

The numbers are (a – d) =

$1\text{ }\text{ }\left( -6 \right)\text{ }=\text{ }1\text{ }+\text{ }6\text{ }=\text{ }7$

$\begin{array}{*{35}{l}} a\text{ }=\text{ }1 \\ \left( a\text{ }+\text{ }d \right)\text{ }=\text{ }1\text{ }+\text{ }\left( -6 \right)\text{ }=\text{ }1\text{ }\text{ }6\text{ }=\text{ }-5 \\ \end{array}$

Therefore, the numbers

$-5,\text{ }1,\text{ }7,\ldots$

and

$7,\text{ }1,\text{ }-5,\ldots$

are in A.P.

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