mathmath333
  • mathmath333
question
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} S=\{1,2,3,4,\cdots , 1000\} \hspace{.33em}\\~\\ \end{align}}\) How many arithmatic progressions can be formed from the elements of \(S\) that start with \(1\) and end with \(1000\) and have at least \(3\) elements.
anonymous
  • anonymous
may u pls help me with my math?
anonymous
  • anonymous
ok, let's work on this problem

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anonymous
  • anonymous
\(a_1=1\), \(a_n=1000\) let difference of terms be \(d\)
mathmath333
  • mathmath333
u mean common difference
anonymous
  • anonymous
yeah
anonymous
  • anonymous
Now we can calculate \(a_n\)
mathmath333
  • mathmath333
\(a_n=1+(n-1)d\)
anonymous
  • anonymous
very good
anonymous
  • anonymous
then, we have\[999=(n-1)d\]so \(d\) must be a divisor of \(999\)
mathmath333
  • mathmath333
yeas
anonymous
  • anonymous
ok, how many values \(d\) can take?
anonymous
  • anonymous
considering\[999=3^3 \times 37\]
anonymous
  • anonymous
@mukushla can u help me pls
anonymous
  • anonymous
@mathmath333 can u pls help me with my question
mathmath333
  • mathmath333
i didnt undestand about the value of d
anonymous
  • anonymous
well\[n-1=\frac{999}{d}\]right?
mathmath333
  • mathmath333
yes
anonymous
  • anonymous
since \(n-1\) is an integer so \(\frac{999}{d}\) must be an integer, right?
mathmath333
  • mathmath333
yes
anonymous
  • anonymous
so d must be a divisor of 999
mathmath333
  • mathmath333
yes
anonymous
  • anonymous
ok, how many positive divisors does 999 have?
mathmath333
  • mathmath333
i forgot how to do it ,do we add the powers of its prime divisors
anonymous
  • anonymous
yeah, that right,\[(3+1)(2+1)=8\]which are\[1, 3, 9, 27, 37, 111, 333, 999\]
anonymous
  • anonymous
how many of them are acceptable considering conditions of problem?
mathmath333
  • mathmath333
i m still in doubt how u got this (3+1)(2+1)
anonymous
  • anonymous
sry, It's (3+1)(1+1)
mathmath333
  • mathmath333
ok np
anonymous
  • anonymous
final answer will be 7, see if you can get this number as your final answer
mathmath333
  • mathmath333
ans given is 7 , but how u got thaat
anonymous
  • anonymous
well all of values for \(d\) are acceptable, except \(d=1\) which gives a value of \(n=2\), and our arithmatic progressions have at least 3 elements
anonymous
  • anonymous
for example when you let \(d=333\) then \(n\) becomes:\[n-1=\frac{999}{333}=3\]\[n=4\]
anonymous
  • anonymous
which is acceptable
mathmath333
  • mathmath333
u checked all 8 options
anonymous
  • anonymous
no need to check all of them because it's obvious that all of values give a \(n\) that is acceptable
mathmath333
  • mathmath333
ohk thnx briliant/bright
anonymous
  • anonymous
np, anytime math

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