The third and eleventh term of an arithmetic progression are -15 and 25 respectively. Find the:
A.common difference
B.first term
C.number of terms that will give a sum of 30
Let's solve the problem step-by-step:
Step 1: Find the common difference (d).
The formula for the nth term (Tn) of an arithmetic progression is given by:
Tn = a + (n-1)d
where:
Tn = nth term
a = first term
d = common difference
n = term number
Given that the third term (T3) is -15 and the eleventh term (T11) is 25, we can set up the following equations:
T3 = a + 2d
T11 = a + 10d
Substituting the values we have:
-15 = a + 2d .....(equation 1)
25 = a + 10d .....(equation 2)
Step 2: Solve the simultaneous equations.
We will solve equations 1 and 2 simultaneously to find the values of a and d.
From equation 1, we can express a in terms of d:
a = -15 - 2d .....(equation 3)
Substituting equation 3 into equation 2, we get:
25 = (-15 - 2d) + 10d
Simplifying the equation:
25 = -15 + 8d
Adding 15 to both sides of the equation:
40 = 8d
Dividing both sides by 8, we get:
d = 40/8 = 5
Step 3: Find the first term (a).
Substituting the value of d = 5 into equation 3:
a = -15 - 2(5)
a = -15 - 10
a = -25
Therefore, the first term (a) is -25.
Step 4: Find the number of terms (n) that will give a sum of 30.
The sum of an arithmetic series can be calculated using the formula:
Sn = (n/2) * (2a + (n-1)d)
Given that the sum (Sn) is 30, we can set up the following equation:
30 = (n/2) * (2(-25) + (n-1)(5))
Simplifying the equation:
30 = (-25n + 5(n-1)) / 2
Multiplying both sides by 2:
60 = -25n + 5n - 5
Combining like terms:
60 = -20n - 5
Adding 5 to both sides of the equation:
65 = -20n
Dividing both sides by -20, we get:
n = -65/ -20
n = 13/4
Since the number of terms should be a whole number, we can round up to the nearest whole number:
n ≈ 4
Therefore, the number of terms (n) that will give a sum of 30 is 4.
Summary:
A. The common difference (d) is 5.
B. The first term (a) is -25.
C. The number of terms (n) that will give a sum of 30 is 4.
To find the common difference of an arithmetic progression (AP), we can use the formula:
common difference (d) = (nth term - first term) / (n - 1)
where n represents the position of the term in the AP.
In this case, we are given the third term and the eleventh term:
Third term (a3) = -15
Eleventh term (a11) = 25
To find the common difference:
1. We subtract the third term from the eleventh term:
25 - (-15) = 25 + 15 = 40
2. We find the position of the eleventh term:
n = 11
3. We subtract 1 from the position of the eleventh term:
n - 1 = 11 - 1 = 10
4. We calculate the common difference:
common difference (d) = 40 / 10 = 4
Therefore, the common difference of the arithmetic progression is 4 (A).
To find the first term of the arithmetic progression, we can use the formula:
first term (a) = term - (position - 1) * common difference
Using the values we have:
1. We substitute the value of the eleventh term, first term, and common difference into the formula:
25 = a + (11 - 1) * 4
2. We simplify the equation:
25 = a + 10 * 4
25 = a + 40
3. We isolate the first term:
a = 25 - 40
a = -15
Therefore, the first term of the arithmetic progression is -15 (B).
To find the number of terms (n) that will give a sum of 30, we can use the formula for the sum of an arithmetic series:
sum of series (S) = n/2 * (2a + (n - 1) * d)
Using the values we have:
1. We substitute the value of the first term, common difference, and sum into the formula:
30 = n/2 * (2 * (-15) + (n - 1) * 4)
2. We simplify the equation:
30 = n/2 * (-30 + 4n - 4)
3. We distribute the n/2:
30 = -30n/2 + 2n^2/2 - 2n/2
4. We combine like terms:
30 = -15n + n^2 - n
5. We rearrange the equation in standard form:
n^2 - 17n + 60 = 0
6. We factor the equation:
(n - 5)(n - 12) = 0
7. We solve for n:
n - 5 = 0 or n - 12 = 0
n = 5 or n = 12
Therefore, the number of terms that will give a sum of 30 is either 5 or 12 (C).
a11-a3 = 8d = 40
now find a, and then n such that
n/2 (2a+(n-1d)) = 30