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Delicious77 [7]
3 years ago
13

Prove that the (n+1) term of a G.P of which the first term is a and third term is b,is equal to the (2n+1) term of a G.P of whic

h the first term is a and fifth term is b.
Mathematics
1 answer:
Ahat [919]3 years ago
3 0
Hello,

u_{1} =a\\
 u_{3} =b\\

q= (\dfrac { u_{3} } {u_{1 } }  )^ \frac{1}{2} \\\\\\

=(\dfrac{b}{a}) ^ \frac{1}{2}  \\




 u_{n+1} =a*q^{n}=a* (\dfrac{b}{a}) ^ \dfrac{n}{2} \\


 \\\\\\\\\\\\v_{1} =a\\
 v_{5} =b\\

r= (\dfrac { v_{5} } {v_{1 } }  )^ \frac{1}{4} \\\\\\


 v_{2n+1} =a*r^{2n}\\\\
=a* (\dfrac{b}{a}) ^ \dfrac{2n}{4}\\\\


=a* (\dfrac{b}{a}) ^ \dfrac{n}{2} \\






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Lcm of (x-2)(x+3) and 10(x+3)^2
Arte-miy333 [17]
Least common multiple: factor them, then see what they have in common and what is leftover and multiply those expressions:

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3 years ago
The value of the x-intercept for the graph of 4x-5y=40 is
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\hbox{x-intercept - y=0}: \\
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3 years ago
Dylan's expenditures for each of the past six months were: March $1,248.59; April, $1,365.38; May, $1,024.30; June, $1,100.40; J
ivann1987 [24]

Answer:

The average monthly expenditure is 1,155.35 $.

Step-by-step explanation:

The average of any sequency is given by the sum of it's individual parts divided by the number of parts it has. So in this case the average of monthly expenditure will be the sum of all individual expenditures divided by the number of months. The question can be solved like this:

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3 0
3 years ago
Read 2 more answers
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