The sum of the first 21 terms of the progression,-18,-15,-12,......is:
Ah, the sum of the first 21 terms of that progression? Well, if I could sum it up in one word, it would be "mathematical mischief!" Anyway, let's clown around with this problem, shall we?
Now, if we take a look at the progression -18, -15, -12... it seems like we're just skipping around by 3. That's quite the leap! So, let's find out how far we've come.
The first term is -18 and the common difference is 3. How many leaps (or terms) have we taken? Well, since we're looking for the sum of the first 21 terms, we need to find the 21st term. To calculate that, we use the formula:
nth term = first term + (n - 1) * common difference.
Plugging in the values we know, we get:
21st term = -18 + (21 - 1) * 3.
Now let the numbers do the math for us!
21st term = -18 + 20 * 3.
21st term = -18 + 60.
21st term = 42.
So, the 21st term is 42. And since we're looking for the sum of the first 21 terms, we can use the sum of an arithmetic series formula:
S = (n/2)(2a + (n - 1)d).
Plugging in the values we know, we get:
S = (21/2)(2 * -18 + (21 - 1) * 3).
Now let's calculate the sum of all this clowning around!
S = (21/2)(-36 + 20 * 3).
S = (21/2)(-36 + 60).
S = (21/2)(24).
S = 21 * 12.
S = 252.
So, the sum of the first 21 terms of this mischievous progression is 252. Hope that gives you a mathematical laugh!
To find the sum of the first 21 terms of the arithmetic progression -18, -15, -12, ..., we can use the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is:
Sn = (n/2)(a1 + an)
where Sn represents the sum of the first n terms, a1 is the first term, an is the nth term, and n is the number of terms.
In this case, the first term a1 = -18, the nth term an is unknown, and the number of terms n = 21.
To find the nth term an, we can use the formula for the nth term of an arithmetic sequence:
an = a1 + (n-1)d
where d is the common difference.
Given that the common difference in this arithmetic progression is 3 (since each term is increased by 3), we can substitute the values into the formula:
an = -18 + (21-1)(3)
= -18 + 20(3)
= -18 + 60
= 42
We have found that an = 42, so we can now calculate the sum of the first 21 terms:
Sn = (n/2)(a1 + an)
= (21/2)(-18 + 42)
= (21/2)(24)
= 21 * 12
= 252
Therefore, the sum of the first 21 terms of the progression -18, -15, -12, ... is 252.
To find the sum of the first 21 terms of the given arithmetic progression, we can use the arithmetic series formula.
The arithmetic series formula is represented as:
Sn = (n/2) * (2a + (n-1)d)
Where:
- Sn represents the sum of the first n terms in the arithmetic progression.
- n represents the number of terms in the series.
- a represents the first term of the series.
- d represents the common difference between consecutive terms.
In the given progression, the first term (a) is -18, and the common difference (d) is 3 since each term increases by 3.
Applying the formula, we have:
Sn = (21/2) * [2(-18) + (21-1)3]
= (21/2) * [-36 + 20*3]
= (21/2) * [-36 + 60]
= (21/2) * 24
= 21 * 12
= 252
Therefore, the sum of the first 21 terms of the given progression is 252.
a = -18
d = 3
S21 = 21/2 (2a+20d)
So plug and chug