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

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Question (b)

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

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:

Tn=a+(n1)dT_n = a + (n - 1)d

Where:

  • aa is the first term of the sequence (which is pp in this case),
  • dd is the common difference (which is 77, determined from the sequence),

Thus, substituting the values:

Tn=p+(n1)(7)T_n = p + (n - 1)(7)

Simplifying this gives:

Tn=p+7n7T_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 pp such that 2021 is a term in the sequence, we set:

p+7n7=2021p + 7n - 7 = 2021

Rearranging gives:

p+7n=2028p + 7n = 2028

Thus:

p=20287np = 2028 - 7n

To make sure that pp is a natural number, 20287n2028 - 7n must be greater than or equal to 11:

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.

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