To solve this problem, we need to find the next common number after 19 in the given arithmetic sequences. Let's break down the sequences and identify a systematic approach to find the common number step-by-step.
### First Sequence:
The first sequence starts at 7 and increases by 4 each time. The general term for the first sequence can be written as:
\(\[ a_n = 7 + 4(n-1) \]\)
For n = 1: \(\( a_1 = 7 + 4(1-1) = 7 \)\)
For n = 2: \(\( a_2 = 7 + 4(2-1) = 11 \)\)
For n = 3: \(\( a_3 = 7 + 4(3-1) = 15 \)\)
For n = 4: \(\( a_4 = 7 + 4(4-1) = 19 \)\)
So, the first sequence is: 7, 11, 15, 19, 23, ...
### Second Sequence:
The second sequence starts at 1 and increases by 9 each time. The general term for the second sequence can be written as:
\(\[ b_n = 1 + 9(n-1) \]\)
For n = 1: \(\( b_1 = 1 + 9(1-1) = 1 \)\)
For n = 2: \(\( b_2 = 1 + 9(2-1) = 10 \)\)
For n = 3: \(\( b_3 = 1 + 9(3-1) = 19 \)\)
So, the second sequence is: 1, 10, 19, 28, 37, ...
### Finding the Common Term:
From the given sequences, we see that 19 is a common term (it's present in both sequences).
To find the next common term after 19, we need to look at the subsequent terms in both sequences and identify the next one that matches.
Let's express the sequences starting after 19:
#### First Sequence Terms After 19:
\(\[ a_5 = 19 + 4(1) = 23 \]\)
\(\[ a_6 = 19 + 4(2) = 27 \]\)
\(\[ a_7 = 19 + 4(3) = 31 \]\)
\(\[ a_8 = 19 + 4(4) = 35 \]\)
\(\[ a_9 = 19 + 4(5) = 39 \]\)
\(\[ a_{10} = 19 + 4(6) = 43 \]\)
\(\[ a_{11} = 19 + 4(7) = 47 \]\)
\(\[ a_{12} = 19 + 4(8) = 51 \]\)
#### Second Sequence Terms After 19:
\(\[ b_4 = 19 + 9(1) = 28 \]\)
\(\[ b_5 = 19 + 9(2) = 37 \]\)
\(\[ b_6 = 19 + 9(3) = 46 \]\)
We compare the subsequent terms:
- From first sequence: 23, 27, 31, 35, 39, 43, 47, 51, ...
- From second sequence: 28, 37, 46, 55, 64, ...
By inspecting these, we see that the term 55 is common to both sequences:
\(\[
a_{12} = 19 + 4(9) = 55
\]\)
\(\[
b_7 = 19 + 9(4) = 55
\]\)
### Conclusion:
The next number that is in both sequences after 19 is 55.
