Question

3. The sum of four integers in AP is 24 and their product is 945. Find the numbers,
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Answer 1

Answer:

Let the 4 integers be (a−3d),(a−d),(a+d) and (a+3d) respectively.

Now according to the question-

Sum of these integers is 24.

∴(a−3d)+(a−d)+(a+d)+(a+3d)=24

⇒a−3d+a−d+a+d+a+3d=24

⇒4a=24⇒a=6

Product of these numbers is 945.

∴(a−3d)(a−d)(a+d)(a+3d)=24

⇒(a

2

−(3d)

2

)(a

2

−d

2

)=945

⇒a

4

−10a

2

d

2

+9d

4

=945

Substituting a=6, we have [From(1)]

9d

4

−360d

2

+1296=945

⇒9d

4

−360d

2

+351=0

⇒d

4

−40d

2

+39=0

⇒(d

2

−39)(d

2

−1)=0

⇒d

2

=39,1

Case I:-

d

2

=39

d=

39

=6.24(not possible)

Case II:-

d

2

=1

d=

1

=1

∴a=6 and d=1

Hence the four integers are 3,5,7 and 9