Leaving Cert Mathematics Sequences & Series: p, p + 7, p + 14, p + 21, ... is

p, p + 7, p + 14, p + 21, .. - Leaving Cert Mathematics - Question (b) - 2021

Question (b)

p, p + 7, p + 14, p + 21, ... is an arithmetic sequence, where p ∈ ℕ. (i) Find the nᵗʰ term, Tₙ, in terms of n and p, where n ∈ ℕ. (ii) Find the smallest value of ... show full transcript

Worked Solution & Example Answer:p, p + 7, p + 14, p + 21, .. - Leaving Cert Mathematics - Question (b) - 2021

Step 1

Find the nᵗʰ term, Tₙ, in terms of n and p

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Answer

In an arithmetic sequence, the nᵗʰ term can be determined using the formula:

$T_n = a + (n - 1)d$

Where:

$a$ is the first term of the sequence (which is $p$ in this case),
$d$ is the common difference (which is $7$ , determined from the sequence),

Thus, substituting the values:

$T_n = p + (n - 1)(7)$

Simplifying this gives:

$T_n = p + 7n - 7$

Step 2

Find the smallest value of p for which 2021 is a term in the sequence

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Answer

To find the smallest value of $p$ such that 2021 is a term in the sequence, we set:

$p + 7n - 7 = 2021$

Rearranging gives:

$p + 7n = 2028$

Thus:

$p = 2028 - 7n$

To make sure that $p$ is a natural number, $2028 - 7n$ must be greater than or equal to $1$ :

$2028 - 7n ext{ } egin{cases} ext{ is a natural number} \ ext{ } ext{ is } ext{the nearest multiple of 7 that is less than } 2028. ext{ The closest multiple is } 2023. \ ext{ So:} \ 2028 - p = 2023 \ p = 5 \ ext{Thus, the smallest valid value of } p ext{ is } 5.$

p, p + 7, p + 14, p + 21, .. - Leaving Cert Mathematics - Question (b) - 2021

Worked Solution & Example Answer:p, p + 7, p + 14, p + 21, .. - Leaving Cert Mathematics - Question (b) - 2021

Find the nᵗʰ term, Tₙ, in terms of n and p

Find the smallest value of p for which 2021 is a term in the sequence

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