What is the number of common terms in the two sequences 17, 21, 25, 29

Question

What is the number of common terms in the two sequences 17, 21, 25, 29, ..., 417 and 16, 21, 26, 31, ..., 466?

A. 78
 B. 77
 C. 22
 D. 20
E. 19

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okHedwig · Answer

IMO Option D 
So it took me a long time to understand with what process should I proceed
I solved it like : First common no is 21, next 41 and so on..
so for this series : 21, 41.... a=21, d=20
So the last term should be less or equal to 417
Taking a+(n-1)d <= 417
Solving we get n =20

Let me know in case there is any easier method to solve this.
Thanks!!

himabindureddy · Answer

The sequence 17 - 417 progresses by +4, and the series 16-466 progresses by +5. The common terms 21,41,61..... repeats once in 5 terms in first series. We just need to calculate for the first series as this is the series which ends first(with 417).

There are 100 numbers between 21 and 417 which follow the +4 progression. Hence 100/5 = 20 terms

Ans: D

There are 100 numbers between 21 and 417 which follow the +4 progression. Hence 100/5 = 20 terms

Ans: D

KarishmaB · Answer

17, 21, 25, 29, ..., 417 (each term increases by 4)
16, 21, 26, 31, ..., 466 (each term increases by 5)

First common term = 21 (is there in both sequences)
When we add the LCM of 4 and 5 i.e. 20 or its multiple to 21, the term obtained will belong to both sequences. Hence common terms are of the form 20n + 21. How many such values are less than 417? The smallest term is 21 where n = 0. 
n cannot be 20 because 20*20 + 21 is greater than 417 so last common term will have n = 19.

Total number of such terms = 20

Answer (D)

findingmyself · Answer

To identify the common terms, let's start writing both sequences:

Sq 1: 17,21,25,29,33,37,41.....

Sq2: 16,21,26,31,36,41....

Common terms: 21,41,61,81..401 
 Common diff=20

An=A1+(n-1)d 
 Nth number of a sequence

Here An=401 
 401=21+(N-1)20 
 380=(N-1)20 
 N=19+1=20

Answer D

Another method:

Common difference of two sequences is 4 and 5 
 LCM(4 and 5)=20

Thus 21+(n-1)20<417 
 N-1<396/20<19.6 
 N<20.6 
 Since N is an integer, it should be 20

Ayeka · Answer

17,21,25,29,...,417 (here common difference is 4) 
16,21,26,31,...,466 (here common difference is 5)

Common difference of both sequences = LCM(4,5)= 20 
First common term in both sequences = 21 
Last term must be ≤ min(417, 466) 
Last term = first term + (n-1) common difference 
21+(n-1)20 ≤ 417
(n-1) ≤ (417-21)/20=19.8 
 n ≤ 20.8 
Means n =20

KevoK · Answer

My approach to this:

We can notice that the the 1st sequence has the form of: N1+4, and the 2nd has the form of: N2+5

Because the 1st sequence is progressing slower than the 2nd sequence at the rate of 1 per step, we can conclude that every 5 step of the 1st sequence, or 4 step of the 2nd sequence, there will be one matching element (the sequence of common number is N_s+20).

Testing it out to verify it:

1st sequence : 17,  21,  26, 29, 33, 37,  41,.... 
2nd sequence : 16,  21 , 26, 31, 36,  41,...

IMO, D

Souradeep17 · Answer

series 1: 17,21,25.... +4 inc
series 2: 16,21,26,31..+5 inc

The common terms will be then 21,41 ,61 .. (+20 increase as lcm of 4 & 5 is 20)

ending term is 417 ( considering the lower limit)
 no. of terms
for n=20 we will have 421 , so the number less than 421 is 401. 
(401-21)/20+1=19 +1=20

Total number of terms will be then 19+1 =20 Option D

What is the number of common terms in the two sequences 17, 21, 25, 29

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What is the number of common terms in the two sequences 17, 21, 25, 29

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