The second term of an arithmetic progression is 8 and the eleventh term is -10 Find the first term, the common differrnce and sum to 15 terms
the two terms are 9d apart. So, d = -2
Now you have d, and can easily find a, so
S15 = 15/2 (2a + 14d)
To find the first term (a), the common difference (d), and the sum of the first 15 terms (Sn) of an arithmetic progression, we can use the formulas for the nth term (An) and the sum of the first n terms (Sn) of an arithmetic progression.
The formula for the nth term (An) is:
An = a + (n-1)d
The formula for the sum of the first n terms (Sn) is:
Sn = (n/2)(2a + (n-1)d)
1. Finding the first term (a):
Given that the second term (A2) is 8, we can substitute n = 2 and An = 8 into the formula for the nth term:
8 = a + (2-1)d
8 = a + d
2. Finding the common difference (d):
Given that the eleventh term (A11) is -10, we can substitute n = 11 and An = -10 into the formula for the nth term:
-10 = a + (11-1)d
-10 = a + 10d
Now we have two equations:
a + d = 8 ---- equation (1)
a + 10d = -10 ---- equation (2)
3. Solving the equations:
To solve the simultaneous equations, we can subtract equation (1) from equation (2) to eliminate "a":
(a + 10d) - (a + d) = -10 - 8
a + 10d - a - d = -18
9d = -18
d = -18/9
d = -2
Substituting the value of d into equation (1):
a + (-2) = 8
a - 2 = 8
a = 8 + 2
a = 10
Therefore, the first term is 10 and the common difference is -2.
4. Finding the sum of the first 15 terms (Sn):
Substituting a = 10, d = -2, and n = 15 into the formula for the sum of the first n terms (Sn):
Sn = (15/2)(2(10) + (15-1)(-2))
Sn = 7.5(20 + 14(-2))
Sn = 7.5(20 + (-28))
Sn = 7.5(-8)
Sn = -60
Hence, the sum of the first 15 terms is -60.