THE SUM OF THE FIRST TWO TERM OF A GP IS 6 AND THE SUM OF THE 5TH AND 8TH TERM IS 288. SOLVE
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a+ar = 6
ar^4 + ar^7 = 288
a(1+r) = 6
ar^4(1+r^3) = 288
1+r^3 = (1+r)(1-r+r^2), so
r^4 * a(1+r)(1-r+r^2) = 288
6r^4(1-r+r^2) = 288
r^4(1-r+r^2) = 48
We probably want r an integer, so try to see what factors we have of 48
48 = 1*48
48 = 16*3 - bingo
r = 2
so, a = 2
The sequence is
2,4,8,16,32,64,128,256
a1+a2 = 2+4 = 6
a5+a8 = 32+256 = 288
To solve this problem, we need to find the common ratio and the first term of the geometric progression (GP).
Let's assume the first term of the GP is 'a' and the common ratio is 'r'.
Given that the sum of the first two terms of the GP is 6, we can write the equation as:
a + ar = 6 -- Equation 1
Similarly, the sum of the 5th and 8th term of the GP is 288. We can write the equation as:
ar^4 + ar^7 = 288 -- Equation 2
Now, let's solve these two equations simultaneously:
From Equation 1, we can rewrite it as:
a(1 + r) = 6
Dividing Equation 2 by Equation 1, we get:
(ar^4 + ar^7) / (a(1 + r)) = 288 / 6
r^3 = 48
Taking the cube root of both sides, we find:
r = ∛48
Substituting the value of 'r' into Equation 1, we can solve for 'a':
a(1 + ∛48) = 6
a = 6 / (1 + ∛48)
Therefore, the common ratio 'r' is ∛48 and the first term 'a' is 6 / (1 + ∛48).