The third and fifth term of an arithmetic progression are 10 and-10 respectively.
a)Determine the first and the common difference
t3 = a + 2d = 10
t5 = a + 4d = -10
-a -2d = -10
a + 4d = -10
2d = -20
d = -10
a + 2(-10) = 10
a -20 = 10
a = 30; d = -2
b)The sum of the first 15 terms
I am stuck here for b.
In your last line you have
a = 30, d = -2
but you had found d = -10, which is correct
so a = 30, d = -10
sum(n) = (n/2)(2a + (n-1)d ) , you should know that formula
sum(15) = (15/2)(60 + 14(-10))
= -600
Like it thanks
To find the sum of the first 15 terms of an arithmetic progression, you can use the formula for the sum of an arithmetic series:
Sn = n/2(2a + (n-1)d)
Where:
- Sn is the sum of the first n terms,
- a is the first term,
- d is the common difference, and
- n is the number of terms.
In this case, we know that:
- a = 30 (from part a),
- d = -2 (from part a), and
- n = 15 (we want to find the sum of the first 15 terms).
Plugging these values into the formula, we get:
S15 = 15/2(2(30) + (15-1)(-2))
Simplifying further:
S15 = 15/2(60 + 14(-2))
S15 = 15/2(60 - 28)
S15 = 15/2(32)
S15 = 15/2 * 32/1
S15 = 240
Therefore, the sum of the first 15 terms of the arithmetic progression is 240.
To find the sum of the first 15 terms of an arithmetic progression, you can use the formula:
Sn = (n/2)(2a + (n-1)d)
Where:
- Sn represents the sum of the first n terms
- n represents the number of terms
- a represents the first term
- d represents the common difference
In this case, you know that the first term (a) is 30 and the common difference (d) is -10.
Now, substitute these values into the formula and solve for Sn:
Sn = (15/2)(2(30) + (15-1)(-10))
= (15/2)(60 + 14(-10))
= (15/2)(60 - 140)
= (15/2)(-80)
= 15 * -40
= -600
Therefore, the sum of the first 15 terms of this arithmetic progression is -600.