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