1. Determine whether the sequence is arithmetic, geometric, or neither. Then find the tenth term.?
for the first question you can see that the numerator is just being divided by 2 each time. 64/2=32, 32/2=16, 16/2=8, and so on. If the pattern uses multiplication and division then it is geometric. If the pattern uses addition or subtraction then it is arithmetic. For the 2nd question we have the pattern 1+2=3, 2+3=5, 3+5=8, and so on. Finally, number 8. 150+30+6. If the first n value is 0 then we can write this ∑ (150/(5^n)) =150 + 30 + 6 +1.2 + .... This is geometric so it can be subject to the following equation Sum= (a/(1-r)) So which is variable "a" and which is variable "r"? ∑ (150/(5^n)) = ∑ (a*(r^k) a=150, r=(1/5) Sum=(150/ (1-(1/5)) =375/2 Thus 187.5 Make sure to know Sum= (a/(1-r)) by heart. Are you in Calc two? For me, was the only enjoyable part of Calc 2.
Identify the 10th term for the sequence: 12, 6, 3, 3/2... Write answer as an integer or ratio of two integers.?
aᵪ = a₁rˣ⁻¹ a₁ = 12 a₂ = 12·½, so r = ½ a₁₀ = 12·(½)⁹ = 12/2⁹ = 12/512 = 3/128
Find the 10th term in the geometric sequence –2, 6, –18, ....?
Each term is the previous term multiplied by -3. The nth term is (-1)^n * 2 * 3^(n-1) (-1)^10 * 2 * 3^9 = 39,366 Now, you can solve this with a calculator, as I did, or you can say (-1)^10 is going to be 1, so it's either A or C 3^9 = (3^3)^3 3^3 = 27 The last digit, 7, cubed is 7^3 = 343 3*2 = 6, so the number ends in 6. It must be C.
The seventh and tenth terms of an arithmetic sequence are 45 and 66 respectively. What is the least value of n for which the nth term is greater than 1000?
The seventh and tenth terms of an arithmetic sequence are 45 and 66 respectively. What is the least value of n for which the nth term is greater than 1000?Why would you ask two versions of essentially the same question? (The only difference being 20000 instead of 1000.)You learn partly by seeing an explanation, and partly by doing. Once you have seen the solution to one of these, you should be able to do the other. Hints are better than the complete solution because it makes you learn by doing.Here, to go from the 7th to the 10th term you add the common difference three times, and to go from the first to the 7th you add it six times. Therefore the first term is 45 - 2(66–45) = 3 and the common difference is 21/3 = 7.The sum of the first n terms is the number of terms times the average of the first and the last. Let’s make an approximation. If the first is 3 and n is large the 3 makes little difference, so the sum is about 7n^2/2. So n is about the square root of 285 or about 17. The sum of 17 terms is 17(3+(16*7+3))/2. Work that out and see how many terms you need to add or subtract to get to 1000 or more.
Finding the term number in arithmetic/geometric sequences?
First one is geometric because each term is double the previous. The sequence is a = 1.5*2^n and you start at n=1 (ie n=1 yields 3, n=2 yields 6). Now, this is how I derived that sequence: Obviously, sequences can be written in an infinite number of ways. A simple way is to make your first term occur at n=0. If this happens, then the coefficient is the first number (3 in this case). a = 3*2^n 2 is there because it's the multiplier. remember n=0 in this sequence. Now, this works if you remember n is always one less than the term. But if you want something that starts at n=1, you simply shift the sequence over. a = 3*2^n becomes: a = 3*2^(n-1) a = 3*2^n * 2^(-1) a = 3*2^n * (1/2) a = (3/2) * 2^n a = 1.5 * 2^n plug in 384 and you get 384 = 1.5 * 2^n 384 / 1.5 = 2^n 256 = 2^n log base 2 of each side to get 8 = n Second one is arithmetic because you have constant difference between each value. The constant difference is 6, so that's like your rate of change in a linear equation (y=mx+b is a linear equation). So we have a = 6n+b if you start at n=1, b is -1 (just plug in the 5 for a and 1 for n to get -1 for b). Thus the sequence is: a = 6n-1 plug in 485 to get 485 = 6n-1 486 = 6n 486/6 = n 81 = n My explanation of the first sequence might be a bit advanced, it's something that really clicked for me in calc 2.
What is the 1,000th term in the sequence [math] 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5....[/math]?
Just count terms in triangular numbers, the series of 1+2+3+…infinity. The numbers between the last and current triangular numbers is the number of times that number is in the sequence. Shortcut alert: the total for any continuing series of odd numbers is the total of odd numbers n is derived by n^2. Derive n from an integer by doing (last integer-first integer+2)/2 where you are essentially counting the number between last and first to divide by 2 since they are the difference and how many numbers in between, then 2 for the divided by 2 to make to 1 to count the first digit. The closest integer square less than 1000 is 31, which squares up to 961. So (last integer-first integer(1 in this case)+2)/2=31, last integer+1=62, so 31st term is 61, and there are now 62 62s to 63. That would end 62 at 1023rd term, and 1000th is between 961 and 1023, so 62 is the answer.
Find the 10th term of the sequence: 27,18,12,8,16/3,.......?
27,18,12,8,(16)/(3) This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by (2)/(3) gives the next term. In other words, an=r*an-1. Geometric Sequence: r=(2)/(3) This is the form of a geometric sequence. an=a1r^(n-1) Substitute in the values of a1=27 and r=(2)/(3). an=27*(2)/(3^(n-1)) Multiply 27 by (2)/(3^(n-1)) to get (54)/(3^(n-1)). an=(54)/(3^(n-1)) a = 54/3^(10-1) = 0.002743
How can we find the nth term formula of the sequence "2, 10, 30, 68"?
A2A: How can we find the nth term formula of the sequence "2, 10, 30, 68"?There are infinite many solutions; however, the simplest appears to be[math]\qquad A_{n}=n^{3}+n,\;\forall n\in\left\{ 1,2,3,\dots\right\} .[/math]$ python>>> A = lambda n: n**3 + n>>> [A(n) for n in xrange(1,11)][2, 10, 30, 68, 130, 222, 350, 520, 738, 1010]The question asks how to find the formula. Well, after dividing every entry be 2, you find the formula for the second differences and work your backwards.The second differences can be given by [math]3\left(n-1\right)[/math]. Now find the summation, which will give you the first differences. Then find the summation of that and multiply by 2.Since this appears to be a homework question, I really shouldn't give you any more. So you still have a bit of work to do.
Can someone answer these sequences and series questions for me its algebra 2?
Algebra is constructive, yet you will desire to be waiting to reason those out. a million. the consumer-friendly distinction is the version between one term and the subsequent. It sounds like -2. 2. different than commencing with 3, each and every term has a consumer-friendly distinction of +7. So the 2d term is 10, it extremely is the 1st term plus 7. So the fifteenth term may be 3+ 14(7)= one 0 one. 3. back, the 1st term is two, the consumer-friendly distinction of +3. there's a sprint "trick" right here. in case you have the sum (a million+2+3+4...n), it is comparable to n(n+a million)/2. on your formula S(5) = 5* unique term + consumer-friendly distinction *(a million+2+3+4+5) 4. Geometric sequences has a multiplicative distinction between words. right here, the 1st term is two, the subsequent 2x3, the subsequent 2x3^2, etc. So the 10th term would have the cost 2x3^9 or 2x27x27x27 5. The sequence defined is 5, 15, 40 5, etc as much as the 12 term. different than grinding it out, i don't be attentive to an effortless formula to evaluate the sum.
The first term of an arithmetic sequence is equal to 6 and the common difference is equal to 3. How to find a formula for the nth term and the value of the 50 th term?
For any arithmetic sequence, let the first term be a and the common difference be d.So the first term is a, second terms is a+d, third term is a+2d and so on. So the nth term by observing this pattern must be a+(n-1)dSo we can get the sum of the sequenceS = a + (a+d) + (a+2d)…+ (a+ (n-1)d) where n is the number of termsLet's rewrite the sum in the reverse order asS = (a+(n-1)d) + (a+(n-2)d)+…(a+d) + aOn adding these 2 equations, we get2S = (2a + (n-1)d) + (2a + (n-1)d) + (2a+(n-1)d))… n timesSo 2S = n(2a+(n-1)d)Or S = (n/2) (2a + (n-1)d)Put the required values and you will get your answer