-1
$\begingroup$

What is the sum of first $50$ terms common to the $AP$ $15,19,23,\dots$ and the $AP$ $14,19,24,\dots$? I know that: The common terms start from $19$ and nothing else. I have tried this but I am facing a lot of difficulty. Please help me.

2 Answers 2

1

We have:

  • $a_n=15+\color\red4n$
  • $b_n=14+\color\green5n$

We know that:

  • $LCM(\color\red4,\color\green5)=20$
  • The first common element is $19$

Hence the AP of common elements is $c_n=19+20n$.

And the sum of the first $50$ elements is $50(c_{0}+c_{49})/2=25450$.

  • 0
    Thanks. This is a very good answer. ^_^ – 2017-01-16
  • 0
    @LokeshSangewar: Thanks :) – 2017-01-16
0

Consider that $(4\mathbf{Z}+15) \cap (5\mathbf{Z}+14)=20\mathbf{Z}+19$. Hence $$ \sum_{i=0}^{49}20n+19=20\left(\sum_{i=0}^{49}n\right)+19\cdot 50=20\cdot \frac{49\cdot 50}{2}+19\cdot 50. $$

  • 0
    How did u get that first expression? – 2017-01-16
  • 0
    Ok I got it by thinking a bit! – 2017-01-16
  • 0
    By the Chinese reminder theorem, see here https://en.wikipedia.org/wiki/Chinese_remainder_theorem – 2017-01-16
  • 0
    But why did u take only the sum of first 19 terms? And is 19.20.11 the final answer? – 2017-01-16
  • 0
    Nope, read it now – 2017-01-16
  • 0
    Explain it in simple words please – 2017-01-16
  • 0
    Like u haven't used the mod concept anywhere so how? – 2017-01-16
  • 0
    "The mod concept"? The sum is for the first 50 terms, as you asked in the above. – 2017-01-16
  • 0
    That Chinese remainder theroem is something related to mod and thing right? – 2017-01-16
  • 0
    Well, it is something more general. However, for your exercise it is enough to see that the first airthmetic progression is (a subset of) the set of all integers $x$ such that $x\equiv 3\bmod{4}$. – 2017-01-16