The sum of first four terms is a G.P. is 30 and that of the last four terms is 960. Is its term is 2, find the common ratio.
If the sum of first four terms in a G.P. is 30 and that of the last four terms is 960, then the sum of the first 8 terms is 990
We also know that a = 2
sum of first 4 terms
= 2(r^4 - 1)(r-1) = 30
r^4 - 1 = 15r - 15
r^4 - 15r + 14 = 0
r = 1 works for the equation , but r = 1 would make all the terms of the
GP the same, so clearly not a solution
r^4 - 15r + 14 = 0 , a bit of trial and error ....
(r-1)(r - 2)(r^2 + 3r + 7) = 0
so r = 2 or r = not real
check: sum(4) = 2(2^4 - 1)/(2-1) = 30 , as needed
so r = 2 , BUT it does not work for sum(8)
sum(8) = 2(2^8 - 1)(2-1) = 510 , not 990 as necessary according to your second part of the question.
check your question.
who says there are 8 terms? Now that we have
a=2
r=2
if there are n terms, then we know that
2(2^n-1)/(2-1) - 2(2^(n-4)-1)/(2-1) = 960
2^n - 1 - 2^(n-4) + 1 = 480
2^n - 2^n/16 = 480
15/16 * 2^n = 480
2^n = 512
n = 9
so the sequence is
2, 4, 8, 16, 32, 64, 128, 256, 512
and the sum of the last 4 terms is 960
misread the question.
Interpreted the question as if there were 8 terms, but it simply said that the last 4 terms had a sum of 960
e.g. There could have been 20 terms.
However .....
suppose there were n+4 terms
we found from the first part, that a = 2, and r = 2
then
Sum(last 4 terms) = sum(n+4) - sum(n) = 960
2(2^(n+4) - 1)/(2-1) - 2(2^n - 1)/(2-1) = 960
2^(n+4) - 1 - 2^n + 1 = 480
2^n(2^4 - 1) = 480
2^n = 32
n = 5
So the sequence consists of 9 terms, (not asked for just clearing up my problem) Everything checks out
is there any alternative solution?? except this
To solve this problem, we need to apply the concept of geometric progression (G.P.).
Let's assume the first term of the G.P. is 'a' and the common ratio is 'r'.
We are given that the sum of the first four terms of the G.P. is 30.
Therefore, using the formula for the sum of a G.P., we can write the equation:
a + ar + ar^2 + ar^3 = 30 ---(1)
We are also given that the sum of the last four terms of the G.P. is 960.
Using the same formula, we can write the equation:
ar^(n-3) + ar^(n-2) + ar^(n-1) + ar^n = 960 ---(2)
where n is the term we want to find.
Now, let's substitute the given condition that the term is 2 into equation (2):
a(r^(2-3)) + a(r^(2-2)) + a(r^(2-1)) + ar^2 = 960
ar^-1 + a + ar + ar^2 = 960 ---(3)
To find the common ratio 'r', we can divide equation (3) by equation (1):
(ar^-1 + a + ar + ar^2) / (a + ar + ar^2 + ar^3) = 960 / 30
(ar^-1 + a + ar + ar^2) / 30 = 32
(ar^-1 + a + ar + ar^2) = 960
Now, we need to simplify this expression and solve for 'r'.
Since we know that the term is 2, we can also write equation (1) as:
a + ar + ar^2 + ar^3 = a(1 + r + r^2 + r^3) = 30
Now, dividing equation (1) by a(1 + r + r^2 + r^3) = 30, we get:
1 + r + r^2 + r^3 = 30 / a
Simplifying further, we have:
r^3 + r^2 + r + 1 = 30 / a
r^3 + r^2 + r + 1 - (30 / a) = 0
Now, using this simplified equation, we can find the value of 'r' by using numerical methods or solving it algebraically. However, the exact value for 'r' cannot be determined without knowing the value of 'a' or 'n'.