Question

The sum of three numbers in A.P. is -3, and their product is 8. Find the common difference.

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

The sum of three numbers in A.P. is -3, and their product is 8

TO DETERMINE

THE COMMON DIFFERENCE

EVALUATION

Let

$\: a - d \: \: , \: a \: \: , \: a + d \: \:$

be the three terms of the AP

Now by the condition 1

$a - d + a + a + d = - 3$

$\implies \: 3a = - 3$

$\therefore \: a \: = - 1$

By the condition 2

$(a - d) \times a \times (a + d) = 8$

$\implies \: ( - 1 - d) \times ( - 1) \times ( - 1 + d) = 8$

$\implies \: ( - 1 - d) \times ( - 1 + d) = - 8$

$\implies \: ( 1 - {d}^{2} ) = - 8$

$\implies \: {d}^{2} = 9$

$\implies \: d = \pm \: 3$

If d = 3

$\red{\fbox{The three terms are - 4 , - 1 , 2}}$

If d = - 3

$\red{\fbox{The three terms are 2 , - 1 , - 4}}$

Answer 2

Answer:

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 where, a is the first term and d is the common difference of the A.P

So,

…… (1)

Also,

(Using)

(Using 1)

Further solving for d,

…… (2)

Now, using the values of a and d in the expressions of the three terms, we get,

First term =

So,

Second term = a

So,

Also,

Third term =

So,

Therefore, the three terms are .