what is the fifth term in this math sentence: A (n) = 1/2 + ( n - 1 ) ( 1/4 )?
85542
A(5) = 1/2 + (5-1)(1/4)\nA(5) = 1/2 + (4)(1/4)\nA(5) = 1/2 + 1\nA(5) = 1 1/2
64026
55/5 - 5 = 6\n\nClever, no?\n\n... The answer above only works in one equation if you add the parentheses (5x5 + 5)/5 = 6. Take them out and you have 5x5+5/5 = 25+1 = 26.
96822
E. None of the above.\n\n1/10 + 1/4 + 1/5 + 2/5.\n\nUse 20 as the common denominator.\n\n2/20 + 5/20 + 4/20 + 8/20 = 19/20.\n\n19/20 is not one of the answers.
40442
(2/X+5)+(3/X+4)=(2x/x^2+9x+20)\n\nX = -.341296
1029214
First term = x\nSecond term = 2x\nthird = 2(2x) = 4x\nfourth = 8x\nfifth = 16x\nsixth = 32x\nseventh = 64x\neighth = 128x\n\n8th/5th = 128x/16x = 128/16 = 8/1
238690
Let S=1+2+3+4+5+...+(n-2)+(n-1)+n\nThen it's the same as S=n+(n-1)+(n-2)+...+3+2+1.\nAddding these two formulas you'll get\n2S=(n+1)+(n+1)+(n+1)+...+(n+1).\nSince there are n terms on the right side,\n we conclude that \n2S=n(n+1) or\n\nS=n(n+1)/2.\n\nHence, in your case it will be\nS=1+2+3+...+966=[966*967]/2=467061.
240240
2/a + 8 - 1/5 +20\n\n27 and four fifths plus 2 over "a"
916838
(1/2)(1/2)=1/4\n\nB. 1/4
498351
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.
272670
The first thing to realize is that 1+2+...+k=k(k+1)/2. This means you are adding terms of the form 2/[k(k+1)] where k starts at 2 and goes to n. The next step is to realize that\n2/[k(k+1)]=2/k - 2/(k+1), so you are really adding up\n\n2/2 - 2/3\n+ 2/3 - 2/4\n+ 2/4 - 2/5\n+.....\n+.....\n+ 2/n - 2/(n+1).\n\nAfter cancelling, this gives 2/2-2/(n+1), so the expression for the sum is\n1-2/(n+1)=(n-1)/(n+1).
924311
1.)\nan = a1 * r^(n - 1)\n\na5 = a1 * r^(5 - 1)\n(1/24) = (2/3) * r^4\n(3/24) = 2r^4\n(1/8) = 2r^4\nr^4 = (1/16)\nr = (1/2)\n\nan = (2/3)(1/2)^(n - 1)\n\na1 = (2/3)(1/2)^(1 - 1) = (2/3)(1/2)^0 = (2/3) * 1 = (2/3)\na2 = (2/3)(1/2)^(2 - 1) = (2/3)(1/2)^1 = (2/3)(1/2) = (1/3)\na3 = (2/3)(1/2)^(3 - 1) = (2/3)(1/2)^2 = (2/3)(1/4) = (1/6)\na4 = (2/3)(1/2)^(4 - 1) = (2/3)(1/2)^3 = (2/3)(1/8) = (1/12)\na5 = (2/3)(1/2)^(5 - 1) = (2/3)(1/2)^4 = (2/3)(1/16) = (1/24)\na6 = (2/3)(1/2)^(6 - 1) = (2/3)(1/2)^5 = (2/3)(1/32) = (1/48)\n\na6 = (1/48)\n\n---------------------------------------------\n\n2.)\nan = a1 + (n - 1)d\n\na5 = a1 + (5 - 1)d\n-5 = a1 + 4d\na1 = -4d - 5\n\na7 = a1 + (7 - 1)d\n-2m - 11 = a1 + 6d\na1 = -2m - 11 - 6d\n\n-4d - 5 = -2m - 11 - 6d\n2d = -2m - 6\n2d = 2(-m - 3)\nd = -m - 3\n\na1 = -4(-m - 3) - 5 = 4m + 12 - 5 = 4m + 7\n\na1 = (4m + 7) + (1 - 1)(-m - 3)\na1 = 4m + 7 + 0(-m - 3)\na1 = 4m + 7\n\na2 = 4m + 7 + (2 - 1)(-m - 3)\na2 = 4m + 7 + 1(-m - 3)\na2 = 4m + 7 - m - 3\na2 = 3m + 4\n\na3 = 4m + 7 + (3 - 1)(-m - 3)\na3 = 4m + 7 + 2(-m - 3)\na3 = 4m + 7 - 2m - 6\na3 = 2m + 1\n\na4 = 4m + 7 + 3(-m - 3)\na4 = 4m + 7 - 3m - 9\na4 = m - 2\n\na5 = 4m + 7 + 4(-m - 3)\na5 = 4m + 7 - 4m - 12\na5 = -5\n\na6 = 4m + 7 + 5(-m - 3)\na6 = 4m + 7 - 5m - 15\na6 = -m - 8\n\na7 = 4m + 7 + 6(-m - 3)\na7 = 4m + 7 - 6m - 18\na7 = -2m - 11\n\na8 = 4m + 7 + 7(-m - 3)\na8 = 4m + 7 - 7m - 21\na8 = -3m - 14\n\nANS : a8 = -3m - 14
413999
In step 4 you divided by (a-b) which is zero.\n\nIf you could do this you could just say that 5 times 0 = 8 times 0 means that 5 = 8, by dividing both sides by 0 and canceling it.
921594
(5/6)-(1/2) = (1/3)\n\n(1/3) / (4/9) = (3/4)
238182
p1    p2\n--- = ---\nq1    q2\n\n 5     2\n--- = ---\n 2    q2\n\nCross multiply,\n\n4/5
774202
a) 1/2\nb) 1/4 x 1/4 = 1/8\nc) 1/2 x 1/2 x 1/4 x 1/4 = 1/64\nd) 3/4 x 3/4 x 3/4 x 3/4 x 1/2 x 1/2 x 1/2 x 1/2 = 12/64 = 3/16
650155
simple\nfirst\nwhen we apply ∑ we can write the series as\n(a_2 - a_1) + (a_3 - a_2) + (a_4 - a_3).... (a_n - a_n-1) (a_n+1 - a_n)\n\nwhen u add \nthe terms starting frm a_2(except a_1 & a_n+1) get cancelled with the correseponding terms and u get\na_n+1 + a_1
628975
1, -1, 2, -2, 1/5, -1/5, 2/5, -2/5\n\nThe numerator has to be a divisor of 2, the denominator a divisor of 5.