Find the first two terms of a geometric sequence if t_3+ t_4 = 50 and t_4 + t_5 =100?
t3= a(r^2)
t4= a(r^3)
t5= a(r^4)
a(r^2)+ a(r^4)= 100
-------------------------
a(r^3)+ a(r^4) = 50
ar^3
----- = 2
ar^2
r = 2 <-------
a(r^3)+ a(r^4)= 100
------------------------
a(r^2)+ a(r^3) = 50
ar^4
----- = 2
ar^2
ar^2 = 2
r = 2^(1/2)
ar^2 + ar^3 = 50
ar^2(1+r) = 50 ---- #1
ar^3 + ar^4 = 100
ar^3(1+r) = 100 --- #2
divide #2 by #1
ar^3/ar^2 = 2
r = 2
in #1:
a(4)(3) = 50
12a = 50
a = 50/12 = 25/6
so the first 5 terms are
25/6 , 50/6, 100/6, 200/6, 400/6
check"
3rd + 4th = 50
LS = 100/6 + 200/6
= 300/6 = 50 , yeah
4th + 5th = 100
LS = 200/6 + 400/6 = 100 , yeah
Thank you
To find the first two terms of a geometric sequence, we can use the formulas for the terms of a geometric sequence:
t_n = a * r^(n-1)
where t_n is the nth term of the sequence, a is the first term, r is the common ratio, and n is the position of the term.
Given that t_3 + t_4 = 50 and t_4 + t_5 = 100, we can substitute these equations into the formula:
a * r^2 + a * r^3 = 50 (equation 1)
a * r^3 + a * r^4 = 100 (equation 2)
Now, we can solve these two equations simultaneously to find the values of a and r.
From equation 1, we can factor out a common term of a * r^2:
a * r^2 * (1 + r) = 50
From equation 2, we can factor out a common term of a * r^3:
a * r^3 * (1 + r) = 100
We notice that (1 + r) is a common term in both equations, so we can divide them to eliminate it:
(a * r^2 * (1 + r)) / (a * r^3 * (1 + r)) = 50 / 100
Simplifying, we get:
r^2 / r^3 = 1/2
1/r = 1/2
r = 2
Therefore, we have found the common ratio, which is r = 2.
Now, we can substitute this value back into equation 1 to find the value of a:
a * 2^2 + a * 2^3 = 50
4a + 8a = 50
12a = 50
a = 50/12
a = 25/6
So, we have found the first term a = 25/6 and the common ratio r = 2 of the geometric sequence.
Therefore, the first two terms of the geometric sequence are t_1 = a = 25/6 and t_2 = a * r = (25/6) * 2 = 25/3.