16
As you are taking 3 away each time, the 5th term will be -5.
56
Ok, take the formula dn+(a-d) this is just when having a sequence with a common difference dn+(a-d) when d=common difference, a=the 1st term, n=the nth term - you have the sequence 2, 4, 6, 8... and you want to find the nth term therefore: dn+(a-d) 2n+(2-2) 2n Let's assume you want to find the 5th term (in this case, the following number in the sequence) 2(5) = 10 (so the fifth term is 10)
U5 = a + 5n = 4 U7 = a + 7n = 10 Therefore 2n = 6 and so n = 3 and then a = 4 - 5n = 4 - 15 = -11 So Un = -11 + 3n and therefore, U10 = a + 10n = -11 + 10*3 = -11 + 30 = 19
Each term seems to be double of the previous number starting with 3. Hence 4th term = 24 and 5th is 48
That depends what the pattern of the sequence is.
0.16
As you are taking 3 away each time, the 5th term will be -5.
If you mean: +3 +1 -1 -3 then it is -5
"The recursive form is very useful when there aren't too many terms in the sequence. For instance, it would be fairly easy to find the 5th term of a sequence recursively, but the closed form might be better for the 100th term. On the other hand, finding the closed form can be very difficult, depending on the sequence. With computers or graphing calculators, the 100th term can be found quickly recursively."
Nth number in an arithmetic series equals 'a + nd', where 'a' is the first number, 'n' signifies the Nth number and d is the amount by which each term in the series is incremented. For the 5th term it would be a + 5d
56
1 1 2 3 5 etc start with 1 and 1 and to get the 3rd term 1+1 = 2 add 2nd and 3rd term to get 4th --- 1 + 2 = 3 add 3rd and 4th term to get the 5th --- 2 + 3 = 5 etc
This is arithmetic progression with common difference of minus three...Formula:First Term +[ (number of term you want-1)*(common difference which is negative 3)]ExampleFor the 3RD term: -5=1+[(3-1)*(-3)]=1+[-6]= -5For 5TH term: -11=1+[(5-1)*(-3)]=1+(-12)=-11.: For the 21st term:=1+[(21-1)*(-3)]=1+[-60]= -59:D
Ok, take the formula dn+(a-d) this is just when having a sequence with a common difference dn+(a-d) when d=common difference, a=the 1st term, n=the nth term - you have the sequence 2, 4, 6, 8... and you want to find the nth term therefore: dn+(a-d) 2n+(2-2) 2n Let's assume you want to find the 5th term (in this case, the following number in the sequence) 2(5) = 10 (so the fifth term is 10)
If the sequence is 1,4,10,19,31,...... Then the sequence formula is, 1 + 3/2n(n - 1) Confirm 5th term....1 + (3/2 x 5 x 4) = 1 + 30 = 31 the 6th (next) term = 1 + (3/2 x 6 x 5) = 1 + 45 = 46
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