Posted by Anonymous on Sunday, February 21, 2010 at 8:10am.
three number form an arithmetric sequence. Their sum is 24.
A)If 'a' is the first number and 'd' the common difference, show that a+d=8
B)If the first number is decreased by 1 and the second number by 2, the three numbers then form a geometric sequence. Find the three terms.

maths  Reiny, Sunday, February 21, 2010 at 8:39am
your last 3 postings all deal with the basic definitions of sequences and series and involve translating the English statements into mathematical expressions.
Why don't you show me what you have done for this one so far, and I will gladly help you further.

maths  Anonymous, Sunday, February 21, 2010 at 8:49am
ok cool... a) S/n=n/2 [2a+(n1)d]
24=3/2[2a+2d]
48=3[2a+2d]
16=2a+2d
8=a+d
B)i don't know from here hey...
my name is marco

maths  Reiny, Sunday, February 21, 2010 at 9:02am
ok Marko, it would have been easier to just do this
a + a+d + a+2d = 24
3a + 3d = 24
a + d = 8
b) so now we know that a = 8d
"If the first number is decreased by 1" so term1 = a1
" the second number by 2" , so term2 = a+d  2
our new terms are then
t_{1} = a1 = 8d  1 = 7d
t_{2} = a+d2 = 8dd2 = 6
t_{3} = a+2d = 8d + 2d = 8+d
now these form a GS
then
6/(7d) = (8+d)/6
take over.

correction  maths  Reiny, Sunday, February 21, 2010 at 9:05am
near the end I had a typo but it did not cause an error
t_{2 = a+d2 = 8dd2 = 6 should have been
t2 = a+d2 = 8d+d2 = 6}

maths  Anonymous, Sunday, February 21, 2010 at 9:20am
ok thx.. i got 36=(8+d)(7d)
36= 568d+7dd squared
36=d^2d+56
d^2+d56+36=0
d^2+d20=0
(d4)(d+5)=o
d=4 d=5
therefore
T1=3
T2=6
T3=12
and
T1=12
T2=6
T3=3

maths  Reiny, Sunday, February 21, 2010 at 9:38am
Nice, see how easy they are ?

maths  Anonymous, Sunday, February 21, 2010 at 10:13am
ya thx a ton
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