-2,-4,-6.
37
5n+2 or 5n-2. I'll assume 10n 10,20,30,40,50
To find the first three terms, plug 1, 2, and 3 in for n:Term 1: 2(1) - 1 = 1Term 2: 2(2) - 1 = 3Term 3: 2(3) - 1 = 5
First look for the difference between the terms, for example the sequence: 5, 8, 11, 14... has a difference of 3. This means the sequence follows the 3 times table - i.e. 3n Now since we need the first term to be 5 we add 2 to our rule to make it work. So the nth term of this sequence is 3n + 2.
The sum of a sequence is given by sum = n/2(2a + (n-1)d) where: n = how many a = first number of sequence d = difference between terms of sequence. For the first 22 odd numbers these are: n = 22 a = 1 d = 2 → sum = 22/2(2×1 + (22 - 1)×2)) = 22² = 484 The sum of the first n odd numbers is always n²: sum = n/2(2×1 + (n-1)2) = n/2(1 + (n-1))×2 = n(n) = n²
Which sequence? Oh, that one! The first three terms are 1, 2 and 72.
because you add the first 2 terms and the next tern was the the sum of the first 2 terms.
123456
6
2
3925
a1=2 d=3 an=a1+(n-1)d i.e. 2,5,8,11,14,17....
1, 1 and 2
Sum of 1st 2 terms, A2 = 2 + 4 = 6 Sum of 1st 3 terms, A3 = 2 + 4 + 6 = 12 Sum of 1st 4 terms A4 = 2 + 4 + 6 + 12 = 20 you can create a formula for the sum of the 1st n terms of this sequence Sum of 1st n terms of this sequence = n2 + n so the sum of the first 48 terms of the sequence is 482 + 48 = 2352
Unless otherwise stated any series/sequence starts with n = 1 Then the first three terms are when n = 1, n= 2 and n = 3 and the series formula 4n - 2 gives the related results, 2, 6 and 10.
Ignoring the "9" , then this is a Fibonacci sequence. 2,2,4,6,10 The first two terms are 'seed' terms then successive terms equal the sum of the two previous terms. 2 + 2 = 4 2 + 4 = 6 4 + 6 = 10 The next term would be 6 + 10 = 16.
a, ar, ar^2 and ar^3 where a and r are constants.