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.
Asked
Active
Viewed 102 times
2 Answers
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 $50(c_{0}+c_{49})/2=25450$.
barak manos
- 43,109
-
Thanks. This is a very good answer. ^_^ – Yami Kanashi Jan 16 '17 at 14:56
-
@LokeshSangewar: Thanks :) – barak manos Jan 16 '17 at 14:56
0
Consider that $(4\mathbf{Z}+15) \cap (5\mathbf{Z}+14)=20\mathbf{Z}+19$. Hence $$ \sum_{i=0}^{49}20n+19=20\left(\sum_{i=0}^{49}n\right)+19\cdot 50=20\cdot \frac{49\cdot 50}{2}+19\cdot 50. $$
Paolo Leonetti
- 15,423
- 3
- 24
- 57
-
-
-
By the Chinese reminder theorem, see here https://en.wikipedia.org/wiki/Chinese_remainder_theorem – Paolo Leonetti Jan 16 '17 at 14:46
-
But why did u take only the sum of first 19 terms? And is 19.20.11 the final answer? – Yami Kanashi Jan 16 '17 at 14:49
-
-
-
-
"The mod concept"? The sum is for the first 50 terms, as you asked in the above. – Paolo Leonetti Jan 16 '17 at 14:51
-
That Chinese remainder theroem is something related to mod and thing right? – Yami Kanashi Jan 16 '17 at 14:52
-
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 $x\equiv 3\bmod{4}$. – Paolo Leonetti Jan 16 '17 at 14:53