What is the sum of the first twelve term of an A.P whose 1st term is 15 and the common difference is 13
In Arithmetic Progression:
an = a1 + ( n - 1 ) d
where
a1 = the initial term
d = the common difference of successive members
an = the nth term
In this case:
a1 = 15
d = 13
a12 = a1 + 11 d
a12 = 15 + 11 ∙ 13
a12 = 15 + 143
a12 = 158
The sum of the first n terms of arithmetic progression:
Sn = n ( a1 + an ) / 2
S12 = 12 ∙ ( a1 + a12 ) / 2
S12 = 12 ∙ ( 15 + 158 ) / 2
S12 = 12 ∙ 173 / 2
S12 = 2076 / 2
S12 = 1038
S12 = 12/2 (2*15 + 11*13) = 1038
To find the sum of the first twelve terms of an arithmetic progression (A.P.), you can use the formula:
Sn = (n/2)(2a + (n-1)d)
Where:
Sn = sum of n terms
a = first term
n = number of terms
d = common difference
Given:
a = 15
d = 13
n = 12
Plugging in these values into the formula, we have:
Sn = (12/2)(2(15) + (12-1)(13))
Simplifying the equation further:
Sn = (6)(30 + 11(13))
= 6(30 + 143)
= 6(173)
= 1038
Therefore, the sum of the first twelve terms of the given arithmetic progression is 1038.
To find the sum of the first twelve terms of an arithmetic progression (A.P.) with the first term (a) and common difference (d), you can use the formula:
Sum = n/2 * (2a + (n-1)d)
In this case, the first term (a) is 15, the common difference (d) is 13, and the number of terms (n) is 12.
Now, let's substitute the values into the formula:
Sum = 12/2 * (2 * 15 + (12-1) * 13)
First, we simplify the expression inside the parentheses:
Sum = 6 * (30 + 11 * 13)
Then, calculate 11 * 13:
Sum = 6 * (30 + 143)
Next, find the sum inside the parentheses:
Sum = 6 * 173
Finally, multiply to get the sum:
Sum = 1038
Therefore, the sum of the first twelve terms of the given arithmetic progression is 1038.