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Given that 7, 11, 15, 19, ... is an arithmetic progression, find the (a) tenth term, (b) sum of the first twelve terms.

Mathematics

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Answers

2 parts
a

43\text{43}

b

348\text{348}

Step-by-step solution

7 steps

Given the arithmetic progression 7,11,15,19,…7, 11, 15, 19, \dots.
The first term is a=7a = 7.
The common difference dd is the difference between consecutive terms:
d=11−7=4d = 11 - 7 = 4.

Part a· 3 steps

Find the tenth term.

  1. Use the formula for the nn-th term of an arithmetic progression, Tn=a+(n−1)dT_n = a + (n-1)d.

    For the tenth term, n=10n=10.

    T10=a+(10−1)dT_{10} = a + (10-1)d

  2. Substitute the values of aa, nn, and dd into the formula.

    T10=7+(9)×4T_{10} = 7 + (9) \times 4

  3. Calculate the value.

    T10=7+36T_{10} = 7 + 36 T10=43T_{10} = 43

    The tenth term is 43\boxed{\text{43}}.

Part b

348\text{348}

Find the sum of the first twelve terms.

  1. Use the formula for the sum of the first nn terms of an arithmetic progression, Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d].

    For the sum of the first twelve terms, n=12n=12.

    S12=122[2a+(12−1)d]S_{12} = \frac{12}{2}[2a + (12-1)d]

  2. Substitute the values of aa, nn, and dd into the formula.

    S12=122[2(7)+(11)×4]S_{12} = \frac{12}{2}[2(7) + (11) \times 4]

  3. Simplify the expression inside the brackets.

    S12=6[14+44]S_{12} = 6[14 + 44] S12=6[58]S_{12} = 6[58]

  4. Calculate the sum.

    S12=348S_{12} = 348

    The sum of the first twelve terms is 348\boxed{\text{348}}.

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