Given that the first term of an a.p is 7,and its 10th term is twice the second term.calculate the 19th term, sum of 28thterms,and the difference between the 9thand 6th terms.

1 answer

In an arithmetic progression (A.P.), the \( n \)-th term is given by the formula:

\[ a_n = a + (n - 1)d \]

where:

  • \( a \) is the first term,
  • \( d \) is the common difference,
  • \( n \) is the term number.

Given:

  • The first term \( a = 7 \)
  • The 10th term is twice the second term.

The 10th term ( \( a_{10} \) ) can be calculated as follows:

\[ a_{10} = a + 9d = 7 + 9d \]

The second term ( \( a_2 \) ) is:

\[ a_2 = a + d = 7 + d \]

According to the problem, we have:

\[ a_{10} = 2a_2 \]

Substituting the expressions we have:

\[ 7 + 9d = 2(7 + d) \]

Expanding the right side:

\[ 7 + 9d = 14 + 2d \]

Now, isolate \( d \):

\[ 9d - 2d = 14 - 7 \]

\[ 7d = 7 \]

\[ d = 1 \]

Now that we have \( d = 1 \), we can find the relevant terms:

  1. Calculate the 19th term \( a_{19} \):

\[ a_{19} = a + (19 - 1)d = 7 + 18 \cdot 1 = 7 + 18 = 25 \]

  1. Sum of the 28th terms will be the same as the 28th term because it seems you mentioned "terms" instead of "terms". Thus, we will calculate \( a_{28} \):

\[ a_{28} = a + (28 - 1)d = 7 + 27 \cdot 1 = 7 + 27 = 34 \]

If you are looking for "the sum of the first 28 terms", that would be:

\[ S_n = \frac{n}{2} \cdot (a + a_n) \]

For the first 28 terms:

\[ S_{28} = \frac{28}{2} \cdot (7 + 34) = 14 \cdot 41 = 574 \]

  1. Calculate the difference between the 9th and 6th terms:

\[ a_{9} = a + (9-1)d = 7 + 8 \cdot 1 = 7 + 8 = 15 \]

\[ a_{6} = a + (6-1)d = 7 + 5 \cdot 1 = 7 + 5 = 12 \]

Now, the difference:

\[ a_{9} - a_{6} = 15 - 12 = 3 \]

Summary:

  • The 19th term \( a_{19} \) is 25.
  • The 28th term \( a_{28} \) is 34 (Or the sum of the first 28 terms is 574).
  • The difference between the 9th and 6th terms is 3.