in geometric sequence use the formula for the nth term what is the 24th term? what is the sum of the first 10 terms?
131975
r is 2/1 = 2\n\nformula for nth term in a*r^(n-1)\n\n24th term is 1(2^23) = 8388608\n\nSum of n terms is a(r^n -1)/(r-1)\n\nSum of 10 terms is 1(2^10 -1)/(2-1) = 1023\n\nIn all the formula 'a' is the first term i.e, 1
579245
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3dn (where n=1 to 10)\n\nThere's the first Transition series for you.  It can also be shown as:\n     [Ar],4s2, 3dn
117943
This is a geometric sequence with a common ratio of (-1/4).\n\nLet:\n\nr = common ratio\na1 = 1st term = 80\na2 = 2nd term = -20\na3 = 3rd term = 5\na4 = 4th term = -5/4\na5 = 5th term \na6 = 6th term\na7 = 7th term\n.\n.\n.\na10 = 10th term\n\nSolution:\n\nr = a2/a1 = a3/a2 = a4/a3 = -20/80 = -1/4\n\na1 = 80 Given\na2 = (a1)(r)= 80r = -20 Given\na3 = (a1)(r)(r) = 80r^2 = 5 Given\na4 = (a1)(r)(r)(r) = 80r^3 = -5/4 Given\na5 = (a1)(r)^(5-1) = 80r^4 = 80(-1/4)^4 = 5/16 ANSWER\na6 = (a1)(r)^(6-1) = 80r^5 = 80(-1/4)^5 = -5/64 ANSWER\na7 = (a1)(r)^(7-1) = 80r^6 = 80(-1/4)^6 = 5/256 ANSWER\n.\n.\na10 = (a1)(r)^(10-1) = 80r^9 = 80(-1/4)^9 = -5/16384 ANSWER
63604
In each triple, the difference between the 2nd and 3rd terms is > 0 so \n\n1 + 1/2 - 1/3 + 1/4 + 1/5 -1/6 + 1/7 > 1 + 1/4 + 1/7 + 1/10 + ...\n\nThe nth term of the series on the right is 1/(3n+1) > 1/(3n+3n) = (1/6) x (1/n)  which is going to diverge.
385877
formula=(100+96+92+.........)-(98+94+90+.........)\nsum of arithmetic progression=Sn=(1/2)n[2a+(n-1)d]\n(1/2)*25{[200+24*2]-[196+24*2]}\n(1/2)*25*(248-244)\n(1/2)*25*4\n=50
75669
The sum from n = 0 to infinity of a^n is called a geometric series. It converges if and only if |a| < 1.\n\nIn this case we have the sum from n = 1 to infinity of (x^n)/(9^n), which can also be written as (x/9)^n. So in this case a = x/9, and in order to converge we need |x/9|<1. Thus the interval of convergence is -9 < x < 9.\n\nAs for the sum of the series, well the sum from n = 1 to infinity of a^n is a/(1-a). Here a = x/9, so you've got (x/9)/(1-x/9) = x/(9-x).\n\nBTW, the series is still geometric even if x is not a natural number. Also x needs to be greater than -9.
625932
35\n\nThe sequence is every even-indexed Fibbonacci number plus 1.  So from the original sequence whose value is the sum of the two previous numbers:\n\n0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...\n\nYou take every second number and add 1:\n\n2, 3, 6, 14, 35, 90, 234, ...
228322
U1=4-2*1=4-2=2\nU2=4-2*2=4-4=0\nU3=4-2*3=4-6=-2\nU4=4-2*4=4-8=-4\nU5=4-2*5=4-10=-6\n\nSo, the first five terms are 2,0,-2,-4,-6.\n\nb)There are a few ways of writing this one, but you can simply see that every number is in the form 4n+3 where n goes from 1 to infinity\nso you have 7,11,15,19,23,27,...\nso nth term would be an=4n+3 or an=an-1 + 4\n\nor you can write nth term as 7+4n where n goes from zero (from zero because you must have 7 included) to infinity.
489569
n^2 - 7n - 2\n         *   4n - 5                  Multiply term by term\n----------------\n          -5n^2 + 35n + 10\n4n^3 - 28n^2 - 8n               add these now\n--------------------------------\n4n^3 - 33n^2 + 27n + 10\n\n\n\nOK... it took out the extra spaces so my numbers don't line up anymore :p
518625
okay so what N means is that you are plugging in any integer value for n. integers, as a reminder, a whole, countable numbers like 1, 6, 88, 45798.\nSo teh first four integers (values of n) are obvious 1, 2, 3, 4\nso the sequence is\n(2*1 -5), (2 *2 -5), (2 *3-5), (2 *4-5)\n-3, -1, 1, 3.\n\nHere's one way to prove that it's arithmetic:\nYou can graph them as continuous variables along an x y coordinate system. That is, graph the points (1, -3), (2, -1), (3, 1), and (4, 3). If you connect them, you'll get a straight line. It's no surprise that the equation of that line (if you choose to find it) is y=2x-5. Because the points lie on a straight line, it's arithmetic (as opposed to geometric).
341018
C = (W^2*L^2) / (24T^2)\nC(24T^2) = W^2*L^2\n[C(24T^2)]/L^2 = W^2\nAnswer:\nW = sq.rt {[C(24T^2)]/L^2}
222447
1) Assuming an aritmetic sequence\n\nbn = b1 + (n-1)d \nletting b3=a1 = 20, then b7=a5 = 32, and b25 = b23\na5 = a1 + (5-1)d\n32 = 20 + (5-1)d\n32 = 20 + 4d\n12 = 4d\n3 = d\n\nb25 = a23 = 20 + (23-1)*3\nb25 = 20 + 22*3\nb25 = 20 + 66\nb25 = 86\n=====================================================\n2)\na75 = a1 + (75-1)d; d=-3\n\na75 = 5 + (75-1)*-3\na75 = 5 + 74*-3\na75 = 5 + -222\na75 = -217\n\nSn = n/2(a1+an)\nS75 = 75/2(5 + -217)\nS75 = 75/2(-212)\nS75 = -7950\n===================================================\n3) ????\n\na5 = 5*5 - 4 = 21\na15 = 5*15 - 4 = 71\na25 = 5*25 - 4 = 121\n\nOR MAYBE\n\na1 = 5*1 - 4\na1 = 5 - 4\na1 = 1\na30 = 5*30 - 4\na30 = 150 - 4\na30 = 146\n\nSn = n/2(a1+an)\nS30 = 30/2(1 + 146)\nS30 = 30/2(147)\nS30 = 2205\n================================================\n4)\n\nrth term is nCr-1m^n-(r-1)k^r-1\n\n5th term =\n4C4m^4-(5-1)k^5-1\n4C4 = 1\n\n5th term = k^4
396161
Well let's say the page to which she opened is n.  The seventh page after that is n+7, then n+14, then n+21, all the way up to n+140.  When you add all the n's together, you get 21n, and then there's the sequence of 20 multiples of 7.  So you have:\n21n + 7 + 14 + 21 + ... +140\nSimplified, this is 21n + 1470\nYou know the sum is 1995, so 21n + 1470 = 1995\nSubtract 1470 from both sides, then divide both sides by 21, and you get n = 25.
481214
1 4 2 5 3\n\nBut it's cyclical, so the sequence is 1 4 2 5 3 1 4 2 5 3 1 4 2 5 3 ... and so on.\nYou can start anywhere in the sequence. The first answer you received is correct, but that person chose to start with 3 instead of 1.
356308
180, 294, ...\n\nIt is the place of the number squared times the place of the previous number; (n^2)(n-1)\n\nYou can get the last one I hope!
213741
The common ratio is (1/2)/1 = 1/2\nFormula for sum of terms is a(1-r^n)/(1-r) , where a is the first term, r- common ratio and n is the number of terms.\n\nsum of first ten terms is 1(1-(1/2)^10)/(1-1/2) = 1.9980\n\nsum of first twelve terms is (1- (1/2)^12)/(1-1/2) = 1.9995\n\nif n tends to infinity the sum becomes a/1-r  , in this case 1/(1/2) = 2
383443
26. The pattern has to start 1, 2, 4, 8, 15. I see the differences are 1, 2, 4, 7, and the differences of the differences are 1, 2, 3. The next difference of the difference has to be 4, so, working my way back up, I figure the next difference has to be 11, making the next number 26.\n\nAs for why this makes sense: Consider the question in two dimensions, i.e. into how many parts will many random lines diivide a plane?  That pattern starts 1, 2, 4, 7, 11 - the same numbers on the differences line above.  If you want to get really abstruse, I would submit that the number parts random spaces divide the fourth dimension is 1, 2, 4, 8, 16, 31, 57.  :)\n\n* * * * * *\n\nHey, you really do ask an interesting question, but it appears I'm the only one willing to answer it.  (Yes, I see you asked the same question a few days ago.)  OK, then, let's look at it more systematically:\n\n\nOne-dimensional question: Into how many parts will random points divide a line?\n\nEasy question, obviously:  Zero points leave one part (i.e. the whole line); one point breaks it into two parts, two points into three parts, etc.  The sequence is 1, 2, 3, 4, 5, etc.  Generally, n random points divide a line into n + 1 parts. \n\n\nTwo-dimensional question: Into how many parts will random lines divide a plane?\n\nHarder question, but you can work this one out on paper.  Zero lines leave one part (the whole plane); one line breaks it in two; two lines breaks it in four.  Then it gets tricky - the third line can't go thru all four parts, but it can go through three of them, so you end up with 7 parts.  The fourth line can go through 4 of the 7 parts.  Now, how you'd prove this rigorously I don't know, but the sequence is 1, 2, 4, 7, 11, 16, etc. and is clearly derived from the one-dimensional sequence: start with 1, then add 1 to get 2, then add 2 to get 4, then add 3 to get 7, then add 4 to get 11, etc.  To put this into a formula, n random lines divide a plane into (n^2 + n + 2)/2 parts.        \n\n\nNow for the three-dimensional question: Into how many parts will random planes divide a three-dimensional space?\n\nTo visualize this one, look at a corner of your room.  Obviously, zero planes leave one part (the whole space), one plane  breaks it in two (visualize the floor of your room as the plane, for example), two planes breaks it in four (visualize the floor plus one wall), and three planes breaks it in eight (visualize the floor plus two adjoining walls).  Now, try to imagine the fourth plane - it can't go through all eight parts, but it can go through seven of them, making a total of 15 parts.  The three-dimensional  sequence starts 1, 2, 4, 8, 15, and again we see it is derived from the sequence for the previous dimension: start with 1, then add 1 to get 2, then add 2 to get 4, then add 4 to get 8, then add 7 to get 15, etc.  To get the number for 5 planes, add 11 to get 26.  To put this into a formula, n random planes divide space into (n^3 + 5n + 6)/6 parts.   \n\nBased on what we are seeing, what the heck - why not carry it forward to the fourth dimension.  Using the numbers from the three-dimensional sequence, the sequence becomes 1, 2, 4, 8, 16, 31, 57, etc.  We can conclude, for example, that 6 random three-dimensional spaces divide the fourth dimension in 57 parts.  Now that should really impress your instructor!