Let the nth term of the arithmetic sequence be
t(n)=ax+b
then
t(1)=a(1)+b=7
t(2)=a(2)+b=12
Subtracting, we find
2a-a=12-7 which gives a=5
Substitute back into the formula,
t(1)=5(1)+b=7, so b=2
the complete rule is then
t(n)=5x+2
The nineteenth term is therefore
t(19)=5*19+2=97.
You can try similarly for the other numbers.
Show full steps:
1a) Determine the formula for the nth term for 7,12,17...
b) Determine the value of the 19th term,t19.
2) Determine the number of terms in the sequence 4,12,36...2125764. Show all necessary steps.
There are 15 more questions but I think if you help me with these two questions I might b able to do the rest!
Thanks.
6 answers
what about number 2
Find the rule for number 2, which is a geometric series.
You can tell if it is a geometric series by dividing any term by the previous. If the quotient is the same, then it is a geometric series.
The rule for a geometric sequence is:
t(n)=ar^n
and r is the quotient by dividing any term by the previous, for example, here it is
12/4=3, 36/12=3, ...
so r=3
To find a, you would substitute one of the terms into the rule:
t(1)=a(3)^1=4, so a=4/3
and the complete rule is
t(n)=(4/3)3^n
check:
t(3)=(4/3)(3^3)=(4/3)*27=36... ok.
To find n for t(n)=2125764
equate the rule:
t(n)=(4/3)(3^n)=2125764
isolate unknown
3^n = 2125764*(3/4) = 1594323
Now you need to find the power to which 3 must be raised to get 1594323.
One way to do this is to take log on both sides.
n log(3) = log(1594323)
and on solving for n, we get n = 13.
Substitute into the rule to make sure the answer is correct:
t(13)=(4/3)*(3^13)=2125764 ok.
You can tell if it is a geometric series by dividing any term by the previous. If the quotient is the same, then it is a geometric series.
The rule for a geometric sequence is:
t(n)=ar^n
and r is the quotient by dividing any term by the previous, for example, here it is
12/4=3, 36/12=3, ...
so r=3
To find a, you would substitute one of the terms into the rule:
t(1)=a(3)^1=4, so a=4/3
and the complete rule is
t(n)=(4/3)3^n
check:
t(3)=(4/3)(3^3)=(4/3)*27=36... ok.
To find n for t(n)=2125764
equate the rule:
t(n)=(4/3)(3^n)=2125764
isolate unknown
3^n = 2125764*(3/4) = 1594323
Now you need to find the power to which 3 must be raised to get 1594323.
One way to do this is to take log on both sides.
n log(3) = log(1594323)
and on solving for n, we get n = 13.
Substitute into the rule to make sure the answer is correct:
t(13)=(4/3)*(3^13)=2125764 ok.
we never learnt log..
I did
geometric sequence: a=4 and r=3
tn = a(r)^n-1
2125764 = 4(3)^n-1
531441= 3^n-1
Now I am stuck how do I make 531441 the same as 3 so I can find n..
I did
geometric sequence: a=4 and r=3
tn = a(r)^n-1
2125764 = 4(3)^n-1
531441= 3^n-1
Now I am stuck how do I make 531441 the same as 3 so I can find n..
You can also do it by trial and error.
raise 3 to the power of 2.
If it is smaller than the targe number, then raise it to 4.
If it is smaller, then to 8.
If it is still smaller, then to 16.
If it is bigger, raise it to (8+16)/2=12.
and so on.
raise 3 to the power of 2.
If it is smaller than the targe number, then raise it to 4.
If it is smaller, then to 8.
If it is still smaller, then to 16.
If it is bigger, raise it to (8+16)/2=12.
and so on.
oh I think I got it:
geometric sequence: a=4 and r=3
tn = a(r)^n-1
2125764 = 4(3)^n-1
531441= 3^n-1
3^12 = 3^n-1
12 = n-1
12+1=n
n=13
Is this way also correct? will I get the full marks for this type of question.
geometric sequence: a=4 and r=3
tn = a(r)^n-1
2125764 = 4(3)^n-1
531441= 3^n-1
3^12 = 3^n-1
12 = n-1
12+1=n
n=13
Is this way also correct? will I get the full marks for this type of question.
1 answer