The sum of an infinite g.p. whose common ratio is numerically less the 1 is 32 and sum of the first two is 24.
C. R. = 1/2 and 16,8,4,2... Or -24,12,-6....
To find the sum of an infinite geometric progression (g.p.), we need to first calculate the common ratio (r) and the first term (a).
Given that the sum of the g.p. is 32, we can use the formula for the sum of an infinite g.p.:
S = a / (1 - r)
Since we know that the sum of the first two terms is 24, we can use another formula for the sum of a finite g.p. with a known number of terms:
S2 = a + ar
Using the given information, we can set up two equations:
Equation 1: 32 = a / (1 - r)
Equation 2: 24 = a + ar
To solve for a and r, we can use a method called substitution or elimination:
First, let's solve Equation 2 for a:
24 = a + ar
a = 24 - ar
Substitute this expression for a in Equation 1:
32 = (24 - ar) / (1 - r)
Now we have a single equation with one variable (r). Solving for r will allow us to find the value of a.
To simplify the equation further, let's multiply both sides by (1 - r) to eliminate the fraction:
32(1 - r) = 24 - ar
Next, distribute the multiplication on the left side:
32 - 32r = 24 - ar
Let's isolate the terms with 'r' on one side:
32 - 24 = -ar + 32r
8 = 32r - ar
8 = r(32 - a)
Now, because r ≠ 1 (as stated in the problem), we can divide both sides by (32 - a):
8 / (32 - a) = r
Once we find the value of r, we can substitute it back into Equation 2 to solve for a:
24 = a + ar
Let's substitute the value of r we found earlier:
24 = a + a(8 / (32 - a))
Now, we have an equation with one variable (a). We need to solve it to find the value of a, which we can then use to calculate the common ratio (r).
a/(1-r) = 32
a+ar = 24