612
3 1.5
96 (each number is twice the previous)
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.
You are adding by 12. 12 + 12 = 24 24 + 12 = 36 36 + 12 = 48
612
3 1.5
96 because you are multiplying by 2, so 6 x 2 is 12, 12 x 2 is 24, 24 x 2 is 48, 48 x 2 is 96, etc.
The common ratio is 2.
It is an Arithmetic Progression with a constant difference of 11 and first term 15.
A single number, such as 4642142824816 does not constitute a sequence.
96 (each number is twice the previous)
It could be either. The answer depends on how many terms if any are between 48 and 192.
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.
They are: 12, 24, 36 and 48
You are adding by 12. 12 + 12 = 24 24 + 12 = 36 36 + 12 = 48
What is the 14th term in the arithmetic sequence in which the first is 100 and the common difference is -4? a14= a + 13d = 100 + 13(-4) = 48