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The third term of an Arithmetic Progression (A.P.) is 23 and the sum of the first seven terms is 210. Find the: a. Common difference; b. First term; c. Sum of the first 20 terms.

Mathematics

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Answers

3 parts
a
7
b
9
c
1510

Step-by-step solution

4 steps

Part a· 2 steps

Common difference

  1. Write down the given information using the formula for the nn-th term of an A.P., an=a+(n−1)da_n = a + (n-1)d, and the sum of nn terms, Sn=n2(2a+(n−1)d)S_n = \frac{n}{2}(2a + (n-1)d).

    Given a3=23a_3 = 23:

    a+(3−1)d=23a + (3-1)d = 23 a+2d=23(1)a + 2d = 23 \quad (1)

    Given S7=210S_7 = 210:

    72(2a+(7−1)d)=210\frac{7}{2}(2a + (7-1)d) = 210 72(2a+6d)=210\frac{7}{2}(2a + 6d) = 210 7(a+3d)=2107(a + 3d) = 210 a+3d=2107a + 3d = \frac{210}{7} a+3d=30(2)a + 3d = 30 \quad (2)
  2. Solve the system of linear equations for dd.

    Subtract equation (1) from equation (2):

    (a+3d)−(a+2d)=30−23(a + 3d) - (a + 2d) = 30 - 23 d=7d = 7

    The common difference is 7\boxed{7}.

Part b9

First term

  1. Substitute the value of dd into equation (1) to find aa.

    a+2(7)=23a + 2(7) = 23 a+14=23a + 14 = 23 a=23−14a = 23 - 14 a=9a = 9

    The first term is 9\boxed{9}.

Part c1510

Sum of the first 20 terms

  1. Use the formula for the sum of nn terms with n=20n=20, a=9a=9, and d=7d=7.

    S20=202(2(9)+(20−1)7)S_{20} = \frac{20}{2}(2(9) + (20-1)7) S20=10(18+19×7)S_{20} = 10(18 + 19 \times 7) S20=10(18+133)S_{20} = 10(18 + 133) S20=10(151)S_{20} = 10(151) S20=1510S_{20} = 1510

    The sum of the first 20 terms is 1510\boxed{1510}.

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