Using the geometric sequence of numbers 1,3,9,27...what is r, the ratio between 2 consecutive terms? show work
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a=1\nar=3\ntherefore r=3/a\n         r=3/1\n  therefore common difference r =3
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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
205641
a) is just sum of an infinite geometric sequence with a=180 and r = (1/3) so thats 180/(1-1/3) = 180*(3/2)=270\nbut it also bounces back so 270*2-180 = 540 - 180 = 360\n\nyou guys remembers the zeno's pardox: a person need to walk from A to B.  But in order to do that he needs to walk to C, where C is midpoint of A and B.  But then he will again need to walk to the midpoint of B and C... etc.  It is inappropriate to say that the person does not ever arrive at B.  This is the same case as part a)
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1113213211\n\nAfter starting the sequence with 1, each term in the sequence consists of groups of two numbers based on the previous term - the first being the quantity and the second specifying which digit.
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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
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1) \n\n1+2+3+4+5+....n +(n-1)+(n-2)+...+3+2+1...regrouping\n\n=2(1+2+3+4+5+....(n-1)) +n...regrouping the term in parenthesis\n\n=2( (1+n-1)+(2+n-2)+(3+n-3)...)+n\n=2(n+n+n+...(n-1) terms)+n\n=2(n)(n-1)/2 +n\n=n^2-n+n\n=n^2\n\n\n2)\n 1=10^0 and 11=(10^1+10^0) and 111=(10^2+10^1+10^0)...\n\n(10^n+10^(n-1)+....10^2+10^1+10^0) times itself\n=(10^0+10^2+10^3+....10^n)^2\n=10^2n+2*10^(2n-1)+.....\n\n\n3)\n\nn^2-n=n(n-1)...one of them must be even and is therefore divisible by two.\n\nn^3-n=n(n^2-1)=n(n+1)(n-1)=(n-1)(n)(n+1) or three consecutive integers.\n\nIf the product is three consecutive integers, then one of them must be divisible by three and therefore the product is divisible by three.
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1, 11, 21, 1211, 111221, 312211, 13112221, ....\n\nthe odd terms (reading from left to right) enumerate the number of times the same figure (which will be specified to its right) appears in the last term of the sequence.\nSo there can be at most three consecutive equal figures in a term: this happens when a figure appears a number of times equal to itself in the previous term, and also the consecutive figure (different from it) appears the same number of times. As the next figure must be different, it cannot give the same figure as a fourth after the first three.\nFor the same reason, the figure before a sequence of three equal must denote a figure different from the one which has been repeated three times, otherwise the count of that figure would have been different.
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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.
1026074
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
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use arithmatic progression formula n(2a+(n-1)d)/2\nwhere n=no. of terms. a=value of first term and d=common difference between each consecutive term\nsay fro your first eg. no. of black ball = 1+3+5+...\nn=10 a=1 d=2\nso sum = 10(2+9*2)/2 = 100\nsolve the rest by changing values of n, a and other variables respectively
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1, 2, 3, 6, 7, 14, 21, 42
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The Fibonacci series has a number of interesting properties.\n\nThe ratio of successive terms (term divided by next term) converges on 0.618 (the Golden Ratio). Get a calculator and fill in the gaps below\n\n5/8 = 0.625\n8/13 =\n13/21 =\n21/34 =\n34/55 =\n55/89 =\n\nAND\n\nThe ratio of successive terms (term divided by previous term) converges to 1.618 = the inverse of the Golden Ratio.\n\n8/5 = 1.6\n13/8 = 1.625\n21/13 =\n34/21 =\n55/34 =\n89/55 =\n\nThe numbers appear in nature eg in the arrangement of leaves on a stalk, sunflower heads, pine cones etc.\n\nThe generating formula, each term is the sum of the last two, can be used to generate other series e.g.\n\n1, 3, 4. 7, 11, 18, 29, 47, 76,\n\nCompare this to\n\n1, 2, 3, 5, 8, 13, 21, 34, 55\n\nand notice how the difference between 55 and 65 is 21, the differemce between 34 and 47 is 13, the difference between 21 and 29 is 8 etc.\n\nThe series starting 1, 4 and 1, 5 and 1, 6  show similar properties\n\n1, 6, 7, 13, 20, 33, 53, 86, 139\n1, 5, 6, 11, 17. 28, 45, 73, 118\n1, 4, 5, 9. 14, 23, 37, 60, 97, \n1, 3, 4. 7, 11, 18, 29, 47, 76,\n1. 2. 3, 5, 8, 13, 21, 34, 55\n\nand they all converge on the golden ratio too.\n\nOne can experiment with other generating formulae\n\neg each term = the sum of the 3 previous terms\n\n0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81\n\nor each term is twice the previous term plus the one prior to that\n\n0, 1, 2, 5, 12, 29, 70, 169\n\nand explore the properties of such series. Have fun!
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For arithmetic progressions, the difference between consequtive terms is a constant.  \n\nfor \n1) each will differ by 3\n2) each differs by 4,\n\nthe general form is a0 + (k-1)*d where d is the difference.\n\nBtw, for geometric progressions, it's the ratio between terms that's constant, not the difference.
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1. 1s2 2s2 2p6 3s1\n2. 1s2 2s2 2p6 \n3. 1s2 2s2 2p4\n4. 1s2 2s2 2p6\n5. 1s2 2s2 2p6 3s2 3p6 3d7 4s2
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A finite goemotric series is a series with finite elements in geomentric progression. ratio of 2 successive terms is constant.\n\nfor example\n\n1,3 , 9 .....\n\nthis is denoted by 1st term a and ratio r so nth term ar^n-1\n\na,a2,ar^2 ....,ar^n-1\n\nif r <1 then the terms keep diminisihing\n\nsum upto n terms is a(r^(n+1)-1)/(r-1)\n\nyou can find morte details in
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As the terms you give are clearly an arithmetic progression, with the difference between successive terms incrementing by 2 each time\n\n7 - 4 = 3\n12 - 7 = 5\n19 - 12 = 7\n28 - 19 = 9\n\nIt is clear that the next few terms are 39, 52, 67, 84, 103, 124 etc.\n\nI recognise this as rather similar to 36, 64, 81, 100, 121 because I am familiar with the squares of the positive integers having the property of being in Arithmetic Progression, with the difference incrementing by 2 (adding a constant, c, won't alter that property), so I can see where this is heading, but let's work the simultaneous equations through to prove what my observation and intuition is telling me ...\n\nThere are values of a b c to be found for which, with n increasing by 1 each time\n\na + b + c = 4 (1)\n4a + 2b + c = 7 (2)\n9a + 3b + c =12 (3)\n\nsubtracting (1) from (2)\n\n3a + b = 3 (4)\n\nsubtracting (2) from (3)\n\n5a + b = 5 (5)\n\nSubtracting (4) from (5)\n\n2a = 2\na = 1\n\nThen, substituting in (4),  b = 0 and substituting in (1) c = 3\n\nAnd it can be seen that this is indeed a sequence of the squares of the positive integers, with 3 added to them, each time.\n\nT(n) = n^2 + 3
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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