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What is the sum of the first 11 terms of the arithmetic series in which a1 -12 and d 5?
Sum = n/2(2a + (n-1)d) = 11/2 x (2 x -12 + 10 x 5) = 11/2 x 26 = 143
What is the fraction that is in the lowest terms and the sum of the terms is nineteen?
7/12 is another.
What is the sum of the first five terms of a geometric series with a1 6 and r 13?
The sum of the first five terms of a geometric series can be calculated using the formula ( S_n = a_1 \frac{1 - r^n}{1 - r} ), where ( a_1 ) is the first term, ( r ) is the common ratio, and ( n ) is the number of terms. Here, ( a_1 = 6 ), ( r = 13 ), and ( n = 5 ). Substituting these values into the formula gives: [ S_5 = 6 \frac{1 - 13^5}{1 - 13} = 6 \frac{1 - 371293}{-12} = 6 \cdot \frac{-371292}{-12} = 6 \cdot 30939 = 185634 ] Thus, the sum of the first five terms is 185634.
What is the sum of 36 divided by 39?
0.9231
What is first term of a geometric series is 3 and the sum of the first term and the second term is 15 What is the sum of the first six terms?
In a geometric series, if the first term ( a ) is 3 and the sum of the first and second terms is 15, we can denote the common ratio as ( r ). Therefore, we have ( 3 + 3r = 15 ), which simplifies to ( 3r = 12 ) or ( r = 4 ). The sum of the first six terms can be calculated using the formula ( S_n = a \frac{1 - r^n}{1 - r} ). Substituting ( a = 3 ), ( r = 4 ), and ( n = 6 ), we get ( S_6 = 3 \frac{1 - 4^6}{1 - 4} = 3 \frac{1 - 4096}{-3} = 4095 ).
Find the sum of the first 48 terms of an aritmetic sequance 2 4 6 8?
To find the sum of the first 48 terms of an arithmetic sequence, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an), where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term. In this case, a1 = 2, n = 48, and an = 2 + (48-1)*2 = 96. Plugging these values into the formula, we get: S48 = 48/2 * (2 + 96) = 24 * 98 = 2352. Therefore, the sum of the first 48 terms of the given arithmetic sequence is 2352.
What is the sum of the first 11 terms of the arithmetic series in which a1 -12 and d 5?
Sum = n/2(2a + (n-1)d) = 11/2 x (2 x -12 + 10 x 5) = 11/2 x 26 = 143
How do factor the sum of terms as a product of the greatest common factor and the sum of 72 and 60?
12(6 + 5)
What is the fraction that is in the lowest terms and the sum of the terms is nineteen?
7/12 is another.
What is the sum of the first five terms of a geometric series with a1 6 and r 13?
The sum of the first five terms of a geometric series can be calculated using the formula ( S_n = a_1 \frac{1 - r^n}{1 - r} ), where ( a_1 ) is the first term, ( r ) is the common ratio, and ( n ) is the number of terms. Here, ( a_1 = 6 ), ( r = 13 ), and ( n = 5 ). Substituting these values into the formula gives: [ S_5 = 6 \frac{1 - 13^5}{1 - 13} = 6 \frac{1 - 371293}{-12} = 6 \cdot \frac{-371292}{-12} = 6 \cdot 30939 = 185634 ] Thus, the sum of the first five terms is 185634.
What is the sum of the first 12 terms of the arithmetic sequence?
The sum of the first 12 terms of an arithmetic sequence is: sum = (n/2)(2a + (n - 1)d) = (12/2)(2a + (12 - 1)d) = 6(2a + 11d) = 12a + 66d where a is the first term and d is the common difference.
What is the sum of 36 divided by 39?
0.9231
What two terms when multiplied gives a product of -12 and sum of 13?
-0.86546 and 13.86546 (approx).
What is the sum of z and 12?
The sum of z and 12 can be calculated by adding the value of z to 12. In mathematical terms, the sum of z and 12 is represented as z + 12. This operation involves combining the numerical value of z with 12 to obtain the total sum.
What is first term of a geometric series is 3 and the sum of the first term and the second term is 15 What is the sum of the first six terms?
In a geometric series, if the first term ( a ) is 3 and the sum of the first and second terms is 15, we can denote the common ratio as ( r ). Therefore, we have ( 3 + 3r = 15 ), which simplifies to ( 3r = 12 ) or ( r = 4 ). The sum of the first six terms can be calculated using the formula ( S_n = a \frac{1 - r^n}{1 - r} ). Substituting ( a = 3 ), ( r = 4 ), and ( n = 6 ), we get ( S_6 = 3 \frac{1 - 4^6}{1 - 4} = 3 \frac{1 - 4096}{-3} = 4095 ).
who find the sum of the first 12 terms of the sequence described by the formula:U n = 3n – 8?
i need it nowww
If the sum of an infinite geometric series is 12 and the common ratio is one third then term 1 is what?
Eight. (8)
