Question

Whe terms be ( (a-3 d),(a-d),(a+d) 8(a+3 d) ) Criven: sum of the nos. in AP-24.
( Rightarrow(a-3 d)+(a-alpha)+(a+a)+(a+3 d)=24 )
( Rightarrow a=6 )
8 the product of these nos. is 945 . ( Rightarrow(a-3 d)(a-d)(a+d)(a+3 d)=945 )
as Putting as b we get ( 9 d^{4}-360 d^{2}+351=0 )
Dividing by 9 we get ( d^{4}-40 d^{2}+39=0 )
( d^{4}-39 d^{2}-d^{2}+39=0 )
( left(d^{2}-1right)left(d^{2}-39right)=0 )
( alpha^{2}-1=0 quad ; quad d^{2}-39=0 )
( d=pi i d^{2}=39 Rightarrow d=sqrt{39}=6.244 ) isn ( ^{prime} t ) possible
( Rightarrow quad d=1 )
( operatorname{sot} a=6 quad 8 d=1 )
( therefore ) me reopured nos. are: ( -(6-3),(6-1),(6+1),(6+3) ) ( =3,5,789 )

# The sum of four integers in A.P. is 24, and their product is 945; find them. The end of four integers in A.P. is 24, and their product is

Solution