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Q: What are the first five terms of the sequence whose nth term is N4 plus 225?

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5, 8, 11, 14 and 17.

5, 7, 9, 11 and 13

4, 6, 8, 10, 12

In simplest terms, 35/16 or 23/16

First, we'll rearrange the expression in the standard format of ax2 + bx + c = 0: 3x2 - 5x = 2 3x2 - 5x - 2 = 0 Now we need to find two numbers whose sum is negative five (coefficient of the second term), and whose product is negative 6 (product of the coefficients of the first and last terms). Those two numbers in this case would be negative six and one. We'll break the middle term down into two terms using those coefficients: 3x2 - 6x + x - 2 = 0 Now we factor out a common multiple between our first two and last two terms: 3x(x - 2) + 1(x - 2) = 0 And we can group common terms: (3x + 1)(x - 2) = 0 Giving us the fully factored expression on the left.

Related questions

5, 8, 11, 14 and 17.

no clue

5

37

9, 17, 25, 33, 41

Here are the first five terms of a sequence. 12 19 26 33 40 Find an expression for the nth term of this sequence.

31

10946

It depends on (a) the first five numbers of what and(b) what sort of sequence.ANY 5 numbers can be put into a quartic sequence. So the answer is: every time.

All but John Adams served two terms. The total of the first five was nine terms or 36 years (almost - Washington's first term was about an month short.)

Sn = -8n + 2S0 = -8(0) + 2 = 2S1 = -8(1) + 2 = -6S2 = -8(2) + 2 = -14S3 = -8(3) + 2 = -22S4 = -8(4) + 2 = -30S5 = -8(5) + 2 = -38

E its the first letter of the number sequence one, two three four five six seven Eight

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