# 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

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.