The fifth term of an arithmetic series is 12 and the eighth term is 3. Find the sum of the first 6 terms and also of the first 11 terms.
T5=a+4d=12
T8=a+7d=3
-3d=9
d=-3
answer is a=24
d=-3
6 term=9
11 terms = -6
To find the value of 'a' in the arithmetic series, we can use the equation for the fifth term:
T5 = a + 4d = 12
Substitute the value of 'd' as -3:
a + 4(-3) = 12
a - 12 = 12
a = 12 + 12
a = 24
So, 'a' is equal to 24.
Now, we can find the sum of the first 6 terms of the arithmetic series. Use the formula for the sum of an arithmetic series:
Sn = n/2 * (2a + (n - 1) * d)
Substitute the values:
S6 = 6/2 * (2(24) + (6 - 1)(-3))
S6 = 3 * (48 + 5(-3))
S6 = 3 * (48 - 15)
S6 = 3 * 33
S6 = 99
The sum of the first 6 terms is 99.
Similarly, we can find the sum of the first 11 terms. Use the same formula:
S11 = 11/2 * (2(24) + (11 - 1)(-3))
S11 = 11/2 * (48 + 10(-3))
S11 = 11/2 * (48 - 30)
S11 = 11/2 * 18
S11 = 11 * 9
S11 = 99
The sum of the first 11 terms is also 99.
To find the sum of the first 6 terms, we need to find the value of the first term (a) and the common difference (d) of the arithmetic series.
From the given information, we can set up a system of equations:
T5 = a + 4d = 12 ...(1)
T8 = a + 7d = 3 ...(2)
Subtracting equation (2) from equation (1) eliminates 'a' and gives:
3d = 9
Solving for d, we find that the common difference is d = -3.
Substituting this value back into equation (2), we can solve for 'a':
a + 7(-3) = 3
a - 21 = 3
a = 3 + 21
a = 24
Now that we have the values of 'a' and 'd', we can find the sum of the first 6 terms using the formula for the sum of an arithmetic series:
S6 = (n/2)(2a + (n - 1)d)
Substituting the values into the formula, we get:
S6 = (6/2)(2(24) + (6 - 1)(-3))
= 3(48 + 5(-3))
= 3(48 - 15)
= 3(33)
= 99
Therefore, the sum of the first 6 terms is 99.
To find the sum of the first 11 terms, we can use the same formula:
S11 = (n/2)(2a + (n - 1)d)
Substituting the values into the formula, we get:
S11 = (11/2)(2(24) + (11 - 1)(-3))
= (11/2)(48 + 10(-3))
= (11/2)(48 - 30)
= (11/2)(18)
= 99
Therefore, the sum of the first 11 terms is also 99.