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Sergeu [11.5K]
3 years ago
7

The first term of an arithmetic progression is 4X and the second term is X^2. The common difference is 12. Find the possible val

ues of X and the corresponding values of the third term.
Mathematics
2 answers:
DerKrebs [107]3 years ago
5 0
ARITH. PROGRESSION

1st term  =x,
2nd term =x², So the common difference d = x² - x

at the same time they are giving d=12, then: x² - x = 12
 x² - x + 12 =0, solving it gives: x' =6 & x" = - 2
for x= - 2, then  PA => -8, 4, 16 (d=12) ==> 3rd term = 16
& for x=6, then PA ==> 24, 36. 48, (d=12) etc 3rd term = 48

both are correct
katovenus [111]3 years ago
3 0
4X,X^2,_
Common difference=12
second term - first term=12
X^2-4x=12
subtract 12 from both sides
X^2-4X-12=12-12
X^2-4X-12=0
now its a quadratic equation, so use either factor method or quadratic formula,
X^2-6X-2X-12=0
take common
x(X-6)-2(x-6)=0
x-2)(x-6)=0
so x is either 2 or 6 
now use them to find the third term
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