Asked by Anonymous

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.
Answered by Reiny
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.
Answered by Anonymous
ok cool... a) S/n=n/2 [2a+(n-1)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
Answered by Reiny
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 = 8-d

"If the first number is decreased by 1" so term1 = a-1
" the second number by 2" , so term2 = a+d - 2

our new terms are then
t<sub>1</sub> = a-1 = 8-d - 1 = 7-d
t<sub>2</sub> = a+d-2 = 8-d-d-2 = 6
t<sub>3</sub> = a+2d = 8-d + 2d = 8+d

now these form a GS
then

6/(7-d) = (8+d)/6

take over.
Answered by Reiny
near the end I had a typo but it did not cause an error

t<sub>2 = a+d-2 = 8-d-d-2 = 6 should have been

t<sub>2 = a+d-2 = 8-d<b>+d</b>-2 = 6
Answered by Anonymous
ok thx.. i got 36=(8+d)(7-d)
36= 56-8d+7d-d squared
36=-d^2-d+56
d^2+d-56+36=0
d^2+d-20=0
(d-4)(d+5)=o
d=4 d=-5
therefore
T1=3
T2=6
T3=12
and
T1=12
T2=6
T3=3
Answered by Reiny
Nice, see how easy they are ?
Answered by Anonymous
ya thx a ton
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