What is the sum to 15 terms of an arithmetic progression whose 8th term is 4?
W.k.t,
Sn = n/2 {2a + (n-1)d} then
S15 = 15/2 {2a + 14d}
=15/2 {2(a+7d)}
wkt, a+7d = 15
S15 = 15(15)=225
To find the sum of an arithmetic progression, we need to use the formula:
Sum = (n/2)(2a + (n-1)d),
where:
- n is the number of terms
- a is the first term
- d is the common difference
In this case, we are given that the 8th term is 4. Let's determine the values of a and d.
We know that the general formula for an arithmetic progression is:
an = a + (n-1)d,
where:
- an is the nth term
Plugging in the values, we have:
4 = a + (8-1)d
4 = a + 7d
From here, we have two equations:
1. 4 = a + 7d
2. an = a + (n-1)d
We could use these equations to solve for d and a. Alternatively, the easier approach is to use the given information to find the 15th term of the arithmetic progression and then calculate the sum.
We can find the 15th term using equation (2):
a15 = a + (15-1)d
a15 = a + 14d
Now that we have the equation for the 15th term, we can find the sum of the first 15 terms using the summation formula mentioned above:
Sum = (n/2)(2a + (n-1)d)
Sum = (15/2)(2a + (15-1)d)
Sum = 15(a + 7d)
Using equation (1), we can substitute the value of a + 7d:
Sum = 15(4)
Sum = 60
Therefore, the sum of the first 15 terms of the arithmetic progression is 60.