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The sixth term and the eleventh term of a arithmetic progression are 30 and 55 respectively. Find the twenty-first term of the series.a.92 ½b.110c.105d.88 1/3

Question

The sixth term and the eleventh term of an arithmetic progression are 30 and 55 respectively.

Find the twenty-first term of the series.

Options:

  • a. 92 ½
  • b. 110
  • c. 105
  • d. 88 1/3
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Solution

Sure, let's solve this step by step.

  1. In an arithmetic progression, the difference between any two consecutive terms is constant. This difference is called the common difference (d).

  2. We know that the nth term of an arithmetic progression can be found using the formula: a + (n-1)d, where a is the first term and n is the term number.

  3. We are given that the 6th term is 30 and the 11th term is 55. We can use these two pieces of information to find the common difference.

    55 = a + 10d (This is from the 11th term) 30 = a + 5d (This is from the 6th term)

  4. Subtract the second equation from the first to find d: 55 - 30 = (a + 10d) - (a + 5d) 25 = 5d d = 5

  5. Now that we have the common difference, we can find the first term (a) by substituting d = 5 into the second equation: 30 = a + 5*5 30 = a + 25 a = 5

  6. Now we can find the 21st term using the formula a + (n-1)d: 21st term = a + (21-1)d = 5 + 20*5 = 5 + 100 = 105

So, the 21st term of the series is 105 (option c).

This problem has been solved

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