The sum of the first 21 terms of the progression _18_15_12....is

1 answer

The progression given is an arithmetic progression (AP) with the first term \( a = 18 \) and a common difference \( d = 15 - 18 = -3 \).

To find the sum of the first \( n \) terms of an arithmetic progression, we use the formula:

\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]

Here, we want to find the sum \( S_{21} \) of the first 21 terms, so \( n = 21 \).

Substituting the values into the formula:

  • \( a = 18 \)
  • \( d = -3 \)
  • \( n = 21 \)

So,

\[ S_{21} = \frac{21}{2} \times (2 \times 18 + (21 - 1) \times -3) \]

Calculating \( 2 \times 18 \):

\[ 2 \times 18 = 36 \]

Calculating \( (21 - 1) \times -3 \):

\[ (21 - 1) \times -3 = 20 \times -3 = -60 \]

Now we can substitute these results back into the formula:

\[ S_{21} = \frac{21}{2} \times (36 - 60) \]

Simplifying \( 36 - 60 \):

\[ 36 - 60 = -24 \]

Now substitute that back:

\[ S_{21} = \frac{21}{2} \times -24 \]

Calculating \( \frac{21}{2} \times -24 \):

\[ = \frac{21 \times -24}{2} = \frac{-504}{2} = -252 \]

Therefore, the sum of the first 21 terms of the progression is:

\[ \boxed{-252} \]