Solveeit Logo

Question

Question: The number of common terms of the sequence \[17,21,25,...,417\] and \[16,21,26,...,466\] is 1\. \[...

The number of common terms of the sequence 17,21,25,...,41717,21,25,...,417 and 16,21,26,...,46616,21,26,...,466 is
1. 2121
2. 1919
3. 2020
4. None of these

Explanation

Solution

In a sequence of numbers the first term is referred to as aa, the common difference referred as dd, and the last term is referred to as ll. Here we asked to find the common terms from the two sequences. We first need to find the total number of terms in both the sequences. Then we will equate the general nth{n^{th}} terms of both sequences to find the common terms.

Formula used:
In a sequence, if the first term is aa, the common difference is dd, and the last term is ll then the total number in the sequence can be given as n=lad+1n = \dfrac{{l - a}}{d} + 1.
Then the nth{n^{th}} term can be given as a+(n1)da + (n - 1)d.

Complete step by step answer:
The given two sequences are, 17,21,25,...,41717,21,25,...,417and16,21,26,...,46616,21,26,...,466.
Let us first find the total number of terms for both sequences.
The first sequence:
The given sequence is 17,21,25,...,41717,21,25,...,417. Here the first term a=17a = 17, the common differenced=2117=4d = 21 - 17 = 4, and the last terml=417l = 417.
Thus, the total number of terms in this sequence is n=lad+1n = \dfrac{{l - a}}{d} + 1.
417174+1\Rightarrow \dfrac{{417 - 17}}{4} + 1
On simplifying this, we get
n=4004+1n = \dfrac{{400}}{4} + 1
On further simplification, we get
n=100+1=101n = 100 + 1 = 101
Therefore, the total number of terms in the first sequence n=101n = 101.
The second sequence:
The given sequence is 16,21,26,...,46616,21,26,...,466. Here the first term a=16a = 16, the common differenced=2116=5d = 21 - 16 = 5, and the last term l=466l = 466.
Thus, the total number of terms in this sequence is n=lad+1n = \dfrac{{l - a}}{d} + 1.
466165+1\Rightarrow \dfrac{{466 - 16}}{5} + 1
On simplifying this, we get
n=4505+1n = \dfrac{{450}}{5} + 1
On further simplification, we get
n=90+1=91n = 90 + 1 = 91
Therefore, the total number of terms in the first sequence n=91n = 91.
Now let us assume that the first sequence has a nn number of terms and the second sequence has amm number of terms.
Now the nth{n^{th}}term of the first sequence is given by the formula a+(n1)da + (n - 1)d. Now substitute the values in a+(n1)da + (n - 1)d
17+(n1)4\Rightarrow 17 + (n - 1)4
On simplifying this, we get4n+134n + 13.
Then the mth{m^{th}} term of the second sequence is given by the formula a+(m1)da + (m - 1)d. Now substitute the values in a+(m1)da + (m - 1)d
16+(n1)5\Rightarrow 16 + (n - 1)5
On simplifying this, we get 5m+115m + 11
Then the common terms in both the sequence will be 4n+13=5m+114n + 13 = 5m + 11
On simplifying the above expression, we get
4n+13=5m+114n + 13 = 5m + 11
5m4n=2\Rightarrow 5m - 4n = 2
Let us find the same possible values of the terms m&nm\& n that satisfy the above equation are2,7,12,...,972,7,12,...,97.
(That is, let m&nm\& n be two. Then, 5(2)4(4)=25(2) - 4(4) = 2
Let m&nm\& n be seven. Then, 5(7)4(7)=75(7) - 4(7) = 7and so on)
Now let us find the total number of terms in the sequence 2,7,12,...,972,7,12,...,97.
Here we have a = 2$$$$d = 7 = 2 = 5$$$$\& $$$$l = 97. Then the total number of terms in that sequence isn=lad+1=9725+1n = \dfrac{{l - a}}{d} + 1 = \dfrac{{97 - 2}}{5} + 1.
On simplifying this, we get
n=955+1n = \dfrac{{95}}{5} + 1
On further simplification, we get
n=19+1=20n = 19 + 1 = 20
Thus, the total number of terms in the sequence 2,7,12,...,972,7,12,...,97is 2020
Let us see the options, option (1) 2121 is not the correct answer since we got the answer 2020 in our calculation.
Option (2) 1919 is not the correct answer since we got the answer 2020 in our calculation.
Option (3) 2020 is the correct answer as we got the same answer in our calculation.
Option (4) None of these is an incorrect answer since we got option (3) as a correct answer.
Hence, option (3) 2020 is the correct option.

Note: Here we are supposed to find the common term in both the sequence thus, we have equated {n^{th}}$$$$\& $$$${m^{th}} terms so that it will be the same and equal. Using the formula that we can find the nth{n^{th}} term of both the sequences (general form).