All three series converge, so the answer is D.
The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.
Consider a geometric sequence with the first term <em>a</em> and common ratio |<em>r</em>| < 1. Then the <em>n</em>-th partial sum (the sum of the first <em>n</em> terms) of the sequence is
Multiply both sides by <em>r</em> :
Subtract the latter sum from the first, which eliminates all but the first and last terms:
Solve for 
:
Then as gets arbitrarily large, the term 
will converge to 0, leaving us with
So the given series converge to
(I) -243/(1 + 1/9) = -2187/10
(II) -1.1/(1 + 1/10) = -1
(III) 27/(1 + 1/3) = 18
Answer:age of the man: 75 age of the woman:25
Step-by-step explanation:
First, We need to define the variables
x: age of the man
y: age of the woman
at the first time he has three times her age
x=3y (1)
in 25 years time
(x+25)=2(y+25) (2)
we clear the equation
X+25=2y+50
X=2y+25
we substitute in the (1) equation:
2y+25=3y
y=25
x=3*25=75
Answer:
I am not sure but i think B so two solutions
Answer:
Given - (a-b)²=7
(a²+b²) =29
To find - ab
Solution -
( a- b) ^2 = 7
(a^2 + b^2) = 29
Expanding the bracket by using identity
a^2 + b^2 -2ab = 7
29 - 2ab = 7 (given a^2 + b^2 = 29 )
-2ab = 7 -29
-2ab = (-22)
ab = (-22) /(-2)
ab = 11
