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.
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.
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
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
t1 = a-1 = 8-d - 1 = 7-d
t2 = a+d-2 = 8-d-d-2 = 6
t3 = a+2d = 8-d + 2d = 8+d
now these form a GS
then
6/(7-d) = (8+d)/6
take over.
near the end I had a typo but it did not cause an error
t2 = a+d-2 = 8-d-d-2 = 6 should have been
t2 = a+d-2 = 8-d+d-2 = 6
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