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Given an arithmetic sequence with the first term a = 6 and the 20th term T20 = 82.

Mathematics

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3 parts
a

d=4d = 4

b
402
c

The 56th term\text{The 56th term}

Step-by-step solution

6 steps

Here's how to solve this problem:

Given an arithmetic sequence with the first term a=6a = 6 and the 20th term T20=82T_{20} = 82.
The formula for the nn-th term of an arithmetic sequence is Tn=a+(n−1)dT_n = a + (n-1)d.

Part a· 2 steps

Find the common difference dd.

  1. Substitute the given values into the formula for T20T_{20}.

    T20=a+(20−1)dT_{20} = a + (20-1)d 82=6+19d82 = 6 + 19d
  2. Solve for dd.

    82−6=19d82 - 6 = 19d 76=19d76 = 19d d=7619d = \frac{76}{19} d=4d = 4

    Answer: d=4\boxed{d = 4}

Part b402

Find the 100th term.

  1. Use the first term a=6a=6 and the common difference d=4d=4 to find T100T_{100}.

    T100=a+(100−1)dT_{100} = a + (100-1)d T100=6+(99)4T_{100} = 6 + (99)4
  2. Calculate the value.

    T100=6+396T_{100} = 6 + 396 T100=402T_{100} = 402

    Answer: 402\boxed{402}

Part c

The 56th term\text{The 56th term}

Which term of this sequence equals 226?

  1. Set Tn=226T_n = 226 in the formula Tn=a+(n−1)dT_n = a + (n-1)d.

    226=6+(n−1)4226 = 6 + (n-1)4
  2. Solve for nn.

    226−6=(n−1)4226 - 6 = (n-1)4 220=(n−1)4220 = (n-1)4 2204=n−1\frac{220}{4} = n-1 55=n−155 = n-1 n=55+1n = 55 + 1 n=56n = 56

    Answer: The 56th term\boxed{\text{The 56th term}}

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