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The 13th term of an arithmetic progression is 10 and the 25th term is 20. Calculate: i) the common difference ii) the first term. iii) the 17th term.

Mathematics

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Answers

3 parts
i

56\frac{5}{6}

iii

403\frac{40}{3}

Step-by-step solution

6 steps

Here's how to solve this arithmetic progression problem:

The formula for the nn-th term of an arithmetic progression is an=a1+(n−1)da_n = a_1 + (n-1)d, where a1a_1 is the first term and dd is the common difference.

Given:
a13=10a_{13} = 10
a25=20a_{25} = 20

This gives us two equations:

  1. a1+(13−1)d=10⇒a1+12d=10a_1 + (13-1)d = 10 \Rightarrow a_1 + 12d = 10
  2. a1+(25−1)d=20⇒a1+24d=20a_1 + (25-1)d = 20 \Rightarrow a_1 + 24d = 20

Part i· 2 steps

Calculate the common difference

  1. Subtract equation (1) from equation (2).

    (a1+24d)−(a1+12d)=20−10(a_1 + 24d) - (a_1 + 12d) = 20 - 10 12d=1012d = 10
  2. Solve for dd.

    d=1012d = \frac{10}{12} d=56d = \boxed{\frac{5}{6}}

Part ii0

Calculate the first term

  1. Substitute the value of dd into equation (1).

    a1+12(56)=10a_1 + 12\left(\frac{5}{6}\right) = 10 a1+2×5=10a_1 + 2 \times 5 = 10 a1+10=10a_1 + 10 = 10
  2. Solve for a1a_1.

    a1=10−10a_1 = 10 - 10 a1=0a_1 = \boxed{0}

Part iii

403\frac{40}{3}

Calculate the 17th term

  1. Use the formula for the nn-th term with n=17n=17, a1=0a_1=0, and d=56d=\frac{5}{6}.

    a17=a1+(17−1)da_{17} = a_1 + (17-1)d a17=0+16(56)a_{17} = 0 + 16\left(\frac{5}{6}\right)
  2. Simplify the expression.

    a17=16×56a_{17} = \frac{16 \times 5}{6} a17=806a_{17} = \frac{80}{6} a17=403a_{17} = \boxed{\frac{40}{3}}
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