How do I solve for n in this expression?
551213
x/(1-n)=kn\nx/k=n(1-n)\nx/k=n-n^2\nn^2-n+x/k=0\n\nUse the quadratic formula on n.
487838
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
787052
1, 1, 9, 9, 25\nFor an odd n, the nth term is n^2.\nFor an even n, the nth term is (n-1)^2.
220245
N + (N+4) =24\n2N +4 =24\n2N = 24-4\n2N=20\nN=10\nN+4 = 14\nElvin is 10, Janet is 14 years old.
222065
nCr = n!/(n - r)!r!\n\nWhere\nx! = x(x - 1)(x - 2)(x - 3)... 1\nExample\n7! = 7*6*5*4*3*2*1 = 5040\nx! reads "x factorial"\n\nso,\n5C4 = 5!/(5 - 4)!4!\n= 5*4*3*2*1 / [1!(4*3*2*1)]\n= 120/1(24)\n= 120/24\n= 5
211777
n=3\nl=1\nm sub l =1,0, or -1\nm sub s = + or - 1/2
896751
are you solving for y or n?
115650
--5 2 0 1 -----  F L O W\n--4 2 0 1 -----  B L O W\n--1 0 2 5 -----  W O L F\n-------------------------\n1 0 4 2 7 ----- N O B L E\n\nBut still N and W value are coming same...does this problem have solution. If you have then can you please send the answer to my mail id hssiddhu@yahoo.com
23824
3n - 16 < or equal to 8n + 29\nSubract both sides by 3n.\n-16 < or equal to 5n +29\nSubtract both sides by 29.\n-45 < or equal to 5n\n-9 < or equal to n\nn > or equal to -9
863437
To get N by itself,\n\nsubtract 5 from both sides of the equal sign.\n\nand multiply by 2 on both sides of the equal sign.\n\nI'm not telling you the answer...someone else might, but I won't.\n\nps: it's not 1!!\n\nThe answer really isn't important here. What is important is knowing how to get the answer.\n\nEDIT:\n\nJUST CURIOUS, WHAT THE HECK IS UP WITH THE THUMBS DOWN ON MY ANSWER AND THE ONE BELOW ME?\n\nWAS THAT THE FIRST PERSON THAT ANSWERED REALIZING WHAT AN IDIOT THEY WERE, AND DELETING THEIR QUESTION?\n\nI JUST DON'T UNDERSTAND PEOPLE SOMETIMES.
1142710
Use this formula, you can find millions of them\n\nL x W = 2(L + W)\n\nuse any number for L and solve for W
748360
Regroup and add.  You got the 5n, so you should be able to get the other part!\n\nn + n - 1 + n - 2 + n + 7 + n - 11 = \nn + n + n + n + n - 1 - 2 + 7 - 11 =\n5n - 7\n\nNotice that -1 -2 - 11 = -14, add 7 to it and you get -7\n\nNRT
522619
1 1 1 2\n1 1 2 1\n1 2 1 1\n2 1 1 1\n\nNo other case is possible, So n(A)=4\n\n\n1 1 1 3\n1 1 3 1\n1 3 1 1\n3 1 1 1\n\n1 1 2 2\n1 2 1 2\n1 2 2 1\n\n2 1 1 2\n2 1 2 1\n2 2 1 1\n\nNo other possibility\nn(B)=10
764334
d=4\na=-1\ntn=a+(n-1)d\n=-1+(n-1)4\n=-1+4n-4\n=4n-5
262222
the answer is 96012000\n\nuse the equation \ns= n( 2*a+(n-1)d)/2
504275
1. The equation for the first would be                      \n\nt(n)={2^(5-1)}*{3^(n-1)}   It is a A.P.\n\n2. It is neither
482632
Multiply the last 2 terms together and then add 1.\nIf that is the pattern (looks like it is) then it should be 155, not 153.\nTo find the next 3 terms, just keep going. (You'll need a calculator with lots of digits).\n22 * 155 + 1 = 3411\n155 * 3411 + 1 = 528706\n3411 * 528706 + 1 = 1803416167