Sequence and series

Algebra Level 3

$3,7,11,\ldots,603 \\ 2,9,16,\ldots,709$

Find the number of terms common to the two arithmetic progressions above.

24 23 22 21

3 solutions

Hung Woei Neoh
Jun 5, 2016

Extend the two APs like this:

$3,7,11,15,19,23,\ldots,603\\ 2,9,16,23,\ldots,709$

The first term common to both APs is $23$ , and the common terms for both APs will also form another AP where $d=\text{lcm}(7,4)=28$ , like this:

$23,51,79,\ldots,T_n$

Now, we need to find the number of common terms $n$ , and we know that the last common term, $T_n$ is less than or equal to $603$ , since the first AP ends at $603$

Therefore,

$T_n \leq 603\\ 23 + (n-1)(28) \leq 603\\ 28n-5 \leq 603\\ 28n \leq 608\\ n \leq 21.71\ldots$

Therefore, $n = \boxed{21}$

How do you prove that the common terms form another arithmetic progression?

Alex Gawkins - 5 years ago

It's simple. The next common term after $23$ will be the term where you add the lowest common multiple of $7$ and $4$ , but here's a way to show it:

Extend the second AP beyond $23$ :

$23,30,37,44,51,58,65,\ldots$

which is the same as this:

$23,23+7,23+14,23+21,23+28,23+35,23+42,\ldots$

Now, since we want the term to also show up in the first AP, it means that the added number can be expressed as $4 + 4 + \ldots + 4$ .

Look at the added numbers:

$7=4+3$ . This is not what we want.

$14=4+4+4+2$ . Nope.

$21 = 4+4+4+4+4+1$ . Not yet.

$28=4+4+4+4+4+4+4$ . $28$ can be obtained by adding the number " $4$ " seven times.

Therefore, the next common term will be $23+28=51$ .

Repeat the process for the next term, and you will get the same result: add $28$ to the current common term to get the next common term.

Since you're repeating a process of adding $28$ , it means that the common terms form an arithmetic progression with common difference $d=28$

I do wonder, @Ashish Siva , do you have a better way to show this?

Hung Woei Neoh - 5 years ago

A Former Brilliant Member
Jun 5, 2016

All the terms in the first sequence leave a remainder of 3 when divided by 4. All the terms in the second sequence leave a remainder of 2 when divided by 7. Using the Chinese Remainder Theorem, common terms would leave a remainder of 7 when divided by 28.

A sequence of common terms would hence be as such: $35,63,...,595$

Then, the answer is obviously 21.

Vivek Sethuraman
Jun 4, 2016

Your answer should proceed on this way; First A.P=3, 7, 11 ...603; common difference =4 So, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51……586 Second A.P=2, 9, 16 …709; common difference=7 So, 2, 9,16,23,30,37,44,51…….586 These terms are the first two common terms of these two A.Ps. now, I consider the colored terms as an A.P. this implies, t1=23; d=28 and tn = 586 tn = a+ (n -1)d Hence, 586 = 23 + (n -1) 28 So, n = 21

Sequence and series

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