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The
term in a certain geometric sequence is and the term in the sequence is . What is the term?Offline
Suppose the first term is a. And the common ratio is r.
ar^22=16
ar^27=24
Dividing the equations, you get r^5=24/16 = 3/2
Raising both sides to the fourth power, r^20 = 81/16
Multiply r^20 with, ar^22, you get ar^42=81/16*16 =81
done!
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
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Thank you, bob.
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
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Hi harrychess,
Also:
To find values of other terms just replace the '43' in either equation with the term number.
In the first 800,000 terms I only found five that are integers.
Last edited by phrontister (2014-05-30 11:34:00)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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Part (a): Find the sum
in terms of a and n.Part (b): Find all pairs of positive integers
such that andI know how to do part a ,but I don't know how to do part b. Could someone explain it to me?
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hi harrychess
You can use the formula from part a to make an equation with a and n as unknowns.
n is one factor here so you only need to check the factors of 200 ={1,2,4,5,8,10,16,20,25,40,50,100,200} as possible values for n.
Calculate a by re-arranging your formula. Is a +ve integer ?
You'll find values of n >10 leads to negative a, so only a few to check. I got three solutions.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Penn writes a 2013-term arithmetic sequence of positive integers, and Teller writes a different 2013-term arithmetic sequence of integers. Teller's first term is the negative of Penn's first term. Each then finds the sum of the terms in his sequence. If their sums are equal, then what is the smallest possible value of the first term in Penn's sequence?
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I got 2 solutions Bob. (18,5) and (9,8). Yes, a and n have to be positive integers.
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Hi Bob,
...the factors of 200 ={1,2,4,5,8,10,16,20,25,40,50,100,200}
16 isn't...
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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Senility I'm afraid. I had the correct list and then edited it to include 16. ??? Thanks for spotting it.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi harrychess;
Penn writes a 2013-term arithmetic sequence of positive integers, and Teller writes a different 2013-term arithmetic sequence of integers. Teller's first term is the negative of Penn's first term. Each then finds the sum of the terms in his sequence. If their sums are equal, then what is the smallest possible value of the first term in Penn's sequence?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi harrychess,
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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