The number of common terms to the two sequences 17, 21, 25, .........., 417 and 16, 21, 26, .........., 466 is

Identify Pattern: First, let's identify the pattern in each sequence. The first sequence starts at 17 and increases by 4 each time (17, 21, 25, ...). The second sequence starts at 16 and increases by 5 each time (16, 21, 26, ...). We need to find the common terms in these sequences.Find nth Term Formula: Let's find the nth term formula for each sequence. For the first sequence, the nth term (a_n) can be given by: a_n = a_1 + (n - 1)d where a_1 is the first term and d is the common difference. For the first sequence, a_1 = 17 and d = 4. So, a_n = 17 + (n - 1) * 4Solve for Common Terms: For the second sequence, the nth term (b_n) can be given by: b_n = b_1 + (n - 1)d where b_1 is the first term and d is the common difference. For the second sequence, b_1 = 16 and d = 5. So, b_n = 16 + (n - 1) * 5Correct Mistake: Now, we need to find the common terms, which means we need to solve for n where a_n = b_n. So we set the nth term equations equal to each other: 17 + (n - 1) * 4 = 16 + (n - 1) * 5Solving the equation for n: 17 + 4n - 4 = 16 + 5n - 5 4n + 13 = 5n + 11 n = 13 + 11 n = 2 This is incorrect because we made a mistake in the simplification process. Let's correct it.

Home

Math Problems

Algebra 2

Find the sum of a finite geometric series

Full solution

Q. The number of common terms to the two sequences

17, 21, 25, \ldots, 417

and

16, 21, 26, \ldots, 466

Identify Pattern: First, let's identify the pattern in each sequence. $\newline$ The first sequence starts at $17$ and increases by $4$ each time ( $17$ , $21$ , $25$ , ...). $\newline$ The second sequence starts at $16$ and increases by $5$ each time ( $16$ , $21$ , $26$ , ...). $\newline$ We need to find the common terms in these sequences.
Find nth Term Formula: Let's find the nth term formula for each sequence. $\newline$ For the first sequence, the nth term $a_n$ can be given by: $\newline$ $a_n = a_1 + (n - 1)d$ $\newline$ where $a_1$ is the first term and $d$ is the common difference. $\newline$ For the first sequence, $a_1 = 17$ and $d = 4$ . $\newline$ So, $a_n = 17 + (n - 1) \times 4$
Solve for Common Terms: For the second sequence, the $n$ th term ( $b_n$ ) can be given by: $\newline$ $b_n = b_1 + (n - 1)d$ $\newline$ where $b_1$ is the first term and $d$ is the common difference. $\newline$ For the second sequence, $b_1 = 16$ and $d = 5$ . $\newline$ So, $b_n = 16 + (n - 1) \times 5$
Correct Mistake: Now, we need to find the common terms, which means we need to solve for $n$ where $a_n = b_n$ . So we set the nth term equations equal to each other: $17 + (n - 1) \cdot 4 = 16 + (n - 1) \cdot 5$
Correct Mistake: Now, we need to find the common terms, which means we need to solve for $n$ where $a_n = b_n$ . So we set the nth term equations equal to each other: $17 + (n - 1) \cdot 4 = 16 + (n - 1) \cdot 5$ Solving the equation for $n$ : $17 + 4n - 4 = 16 + 5n - 5$ $4n + 13 = 5n + 11$ $n = 13 + 11$ $n = 2$ This is incorrect because we made a mistake in the simplification process. Let's correct it.