Question

the sum of three consecutive terms of an AP is 18 and their product is 162, find the terms

Answer 1

ANSWER....

Let required no. be (a-d), a, (a+d).

Then,

(a-d)+ a + (a+d) = 18

$= > 3a = 18$

$= > a = \frac{18}{3}$
$= > a = 6$

And,

(a-d).a.(a+d) = 162

$= > a( {a}^{2} - {d}^{2} ) = 162$
$= > 6(36 - {d}^{2} ) = 162$
$= > 216 - 6 {d}^{2} = 162$
$= > 216 - 162 = 6 {d}^{2}$
$= > 54 = 6 {d}^{2}$
$= > {d}^{2} = \frac{54}{6}$
$= {d}^{2} = 9$

$d = \sqrt{9}$

$d = 3$

Terms are 6,9,12...




hope it helps
thanks...

Answer 2

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GIVEN :

The sum of three terms of an AP = 18

The product of them = 162

TO FIND :

The terms.

SOLUTION:

Let the three terms be (a - d) , (a ), (a + d ).

Where a is the first term of an AP

d is the common difference of an AP.

=> a - d + a + a + d = 18

3a = 18

a = 18/3

a = 6

=> ( a - d) × ( a ) × ( a + d ) = 162

a^3 - a (d^2) = 162

(6) ^3 - 6( d^2) = 162

216 - 162 = 6 (d^2)

54 = 6 (d^2)

9 = d^2

d = 3

The three terms are ,

=> 6 - 3 , 6 , 6 + 3

=> 3 , 6 , 9

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