GATE Overflow Test Series | Quantitative Aptitude | Shallow Test 1: 16

Question

The number of common terms in the two sequences: \( 5, 10, 15, 20, 25, \ldots, 1500 \) and \( 7, 14, 21, 28, 35, \ldots, 1400 \) is _________

• A. 40
• B. 20
• C. 42
• D. 35

Comments

Ans: 40

Both are in AP with a common difference of 5 and 7, but visualize them as multiplication tables of 5 and 7. Since they are asking about common elements. If you see clearly, the 7th table's 7*5=35 and 7*10=70 will also be present in the 5 table. Hence, for every 10 values (example 7*1=7 to 7*10=70) of the 7 table, we are getting two common values.

since tables of 7 end at value 1400(7*200)
200/10=20 (divided by 10 because for every 10 values we get two common)
Total common values = 2*20=40
 

Answer 1

To find the common terms in the two sequences, we need to determine the terms that are present in both arithmetic sequences.

1. First sequence: \( 5, 10, 15, 20, 25, \ldots, 1500 \)
   - This is an arithmetic sequence with the first term \( a_1 = 5 \) and common difference \( d_1 = 5 \).
   - The \( n \)-th term of this sequence is given by:
     \[ a_n = 5 + (n-1) \cdot 5 = 5n \]
   - The sequence ends at 1500, so:
     \[ 5n = 1500 \]
     \[ n = \frac{1500}{5} = 300 \]

2. Second sequence: \( 7, 14, 21, 28, 35, \ldots, 1400 \)
   - This is an arithmetic sequence with the first term \( b_1 = 7 \) and common difference \( d_2 = 7 \).
   - The \( m \)-th term of this sequence is given by:
     \[ b_m = 7 + (m-1) \cdot 7 = 7m \]
   - The sequence ends at 1400, so:
     \[ 7m = 1400 \]
     \[ m = \frac{1400}{7} = 200 \]

The common terms in both sequences will be the terms that are multiples of both 5 and 7. These are the terms that are multiples of the least common multiple (LCM) of 5 and 7.

\[ \text{LCM}(5, 7) = 35 \]

Now, we need to find the multiples of 35 within the range of both sequences:

- The first sequence ranges from 5 to 1500.
- The second sequence ranges from 7 to 1400.

The common multiples of 35 will be:
\[ 35, 70, 105, 140, \ldots \]

The general form of these common multiples is:
\[ 35k \]

For the first sequence:
\[ 35k \leq 1500 \]
\[ k \leq \frac{1500}{35} \approx 42.857 \]
So, the maximum integer value for \( k \) is 42.

For the second sequence:
\[ 35k \leq 1400 \]
\[ k \leq \frac{1400}{35} = 40 \]
So, the maximum integer value for \( k \) is 40.

Thus, the number of common terms is determined by the smallest upper bound of \( k \):
\[ k = 40 \]

So, there are 40 common terms in both sequences.

Correct Answer: A. 40

Comments

Common terms must be numbers that are multiples of both 5 and 7, i.e. multiples of LCM(5,7) = 35.

So we count multiples of 35 that lie in both sequences.

• First sequence goes up to 1500
• Second sequence goes up to 1400 → this is the limiting upper bound

So we count multiples of 35 up to 1400:

1400/35=40A

Therefore, the number of common terms is 40.