Question

Find three consecutive terms in AP whose sum is 54 and the product is 5382.​

Answer 1

a-d+a+a+d = 54

3a = 54

a = 18

(a-d)(a)(a+d) = 5382

(a^2-d^2)(a) = 5382

a^3 - d^2 a = 5382

5832 - 18d^2 = 5382

18d^2 = 450

d^2 = 25

d = ± 5

When d = 5

a-d = 13

a = 18

a+d = 23

When d = -5

a-d = 23

a = 18

a+d = 13

Answer 2

The three terms of A.P are 13,18 and 23.

Step-by-step explanation:

To find : Three consecutive terms in AP whose sum is 54 and the product is 5382 ?

Solution :

Let the three consecutive terms in A.P are (a-d), a, (a+d).

The sum of three consecutive terms in AP is 54.

i.e. $a-d+a+a+d=54$

$3a=54$

$a=\frac{54}{3}$

$a=18$

The product of three consecutive terms in AP is 5382.

i.e. $(a-d)(a)(a+d)=5382$

$(a^2-d^2)(a)=5382$

Substitute the value of a,

$(18^2-d^2)(18)=5382$

$324-d^2=\frac{5382}{18}$

$324-d^2=299$

$d^2=324-299$

$d^2=25$

$d=\pm 5$

When a=18 and d=5,

a-d=18-5=13

a=18

a+d=18+5=23

The three consecutive terms of AP are 13,18,23.

When a=18 and d=-5,

a-d=18-(-5)=23

a=18

a+d=18+(-5)=13

The three consecutive terms of AP are 23,18,13.

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