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Sequence and Series

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Question : 15 of 86

Marks: +1, -0

Sum of three numbers in A.P. is 12 and the sum of their cube is

408 .

The numbers are

$3,4,5$
$1,4,7$
$2,4,6$
None

Solution:

Concept:

Represent three numbers in A.P. as

a-d

,

a

,

a+d

.

Explanation:

Sum:

(a-d)+a+(a+d)=3a=12 \Rightarrow a=4

.

Numbers:

4-d

,

4

,

4+d

.

Sum of cubes:

(4-d)^3+4^3+(4+d)^3=408

.

Compute:

(4-d)^3=64-48d+12d^2-d^3

,

(4+d)^3=64+48d+12d^2+d^3

.

Sum:

192+24d^2=408 \Rightarrow 24d^2=216 \Rightarrow d^2=9 \Rightarrow d=\pm3

.

Numbers:

4-3=1

,

4

,

4+3=7

(or

7,4,1

).

Answer:

1,4,7

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