if the ratio of the sum of n terms of 2 different AP's is (n+1):(n+3) find the ratio of their 6th terms.?
609611
Dear amit.wizkid, thou little knoweth what an abysmal error thou maketh.\n\ntut_einstein, you are better off noting that the sum of the first 'n' terms of an AP, whose first term is 'a' and the difference between terms is 'd' is :\n\nS = (n/2)[2a + (n-1)d]\n\nYou state that \nS1/S2 = (n+1)/(n+3)\n\nwhich is \n\n[2a1 + (n-1)d1] :[2a2 + (n-1)d2] = (n+1) : (n+3) ..... (I)\n\nRemember that what you desire to know is\n(a1 + 5d1) : (a2 + 5d2) \n\nPut n = 11 in (I)\n\n[2a1 + 10 d1] :[2a2 + 10 d2] = 12:14\n\nwhich gives\n\n[a1 + 5d1] : [a2 + 5d2] = 6:7\n\nwhich is what you desire. \n\nShould suffice.
823408
Your ratio is 1:8.15          You fid this out by starting at 13:106 and work yoiur way down to 1. \n\nyou divide both sides by 13 to get your answer.
893261
15a = 3(3a+8)\n15a=9a+24\n15a-9a=24\n6a=24\na=4
390719
so you have (1/2)a+a-(6/3)=4+(1/3)a\n\nadding like terms: (3/2)a-2=4+(1/3)a\n\nsubtract (1/3)a from (3/2)a and add two to both sides:\n(3/2-1/3)a=4+2\n(9/6-2/6)a=6\n(7/6)a=6\n\nnow solve and what do you get? :)
887444
first term a  = 8\n\ncommon ratio r = -1/2\n\nsum of infinite number of terms = a / (1-r)\n\n                             = 8 / [1+(1/2) ]\n\n                              = 16/3
339354
(2x + b) * (a + 3c)\n\n\nsteps:\n2ax + 6xc + ba + 3bc \n1) group first two, and last two\n(2ax + 6xc) + (ba + 3bc)\n\n2) factor out 2x from the first, and then b from the second group\n2x*(a + 3c) +  b*(a + 3c)\n\n3) since (a + 3c) is common, combine 2x and b\n(2x + b)*(a + 3c)
711872
let the number be n\n\nreciprocal of n = 1/n\n\n6 less than n = n-6\n\ntherefore we have:\n(1/n) + [1/(n-6)] = 7/n\n6/n = 1/(n-6)\n6(n-6)  = n\n6n-36 = n\n5n = 36\nn = 36/5
1075653
Here you go! I hope this helps you out. I copied what some other guy said, he was mean about it.  :)\n\nArt B = 2x3 = 6 points\nHist A = 3 x4 = 12 points\nSci C = 4x2 = 8 points\nMath B = 3x3 = 9 points\nLab A = 1 x 4 = 4 points\n\nTotal points = 39\nTotal credits 13\n39/13=3.0\nGPA = 3.0
829616
First, factor the numerator:\nx + 2x² + 3x - 6 --- Combine like terms...\n2x² + 4x - 6 = (2x - 2)(x + 3) = 2(x -1)(x + 3)\n\nNow, you can cancel out the (x - 1) from the numerater and the denominator...\n2(x + 3) / 1 = 2(x + 3)\n\n2(x + 3) is the answer...
1083459
Use this formula:\n\nSn=n/2[t1+tn]\n\nwhere\nSn --> ur required sum\nn --> no. of terms in the given progression \nt1 --> first term (in ur case: 3)\ntn -->last term (in ur case: 803)
1042657
I think you mean 4x/((x+6)(x+6)) + 3/(x-6).  You need a common denominator.  In this case it will be (x+6)(x+6)(x-6).\n\nSo we get 4x(x-6) + 3(x+6)(x+6) as the new numerator.\nThis equals 4x^2-24x + 3x^2+36x+108 = 7x^2+12x+108.\n\nFinal answer: (7x^2+12x+108) / ((x+6)(x+6)(x-6))
1000586
# 1    \n(x+1)*(x+6)=x2 +x+6x+6=x2 +7x +6\n#2\n(y-6)*(y-5)= y2-5y-6y+30= y2-11y+30\n#3\n(a-2) * (a+10)=a2 +10a-2a -20= a2  +8a -20\n#4\n(x-1) *(x+1)= x2 -x+x -1= x2 - 1
51849
9,2,2\nbcoz 6+6+1=9+2+2=13\nThis is the only conflicting triple. as the eldest is one  so their age is 9,2,2
337108
6b + 24a + 45\n= 3*2*b + 3*8*a + 3*15\n= 3(2b + 8a + 15)
889429
1) Yup. a_1 - 5 = a_2, so a_1 must be 3.\n2) What does the common ratio actually mean? Its any term divided by the term before it. So in this case, the common ratio is just (-3)/(-2) = 1.5. Now, how would you get the first term? To go from the first term to the second you'd multiply by 1.5; thus to go from the second to the first, divide by 1.5: -2/1.5 = -4/3 is the first term.
338536
3(6) + 7 = 24\n18 + 7 = 24\n\n25 = 24\n\nNo.
610613
In 6ab + 4a + 3b + 2, we group the first 2 terms and the last 2 terms:\n\n6ab + 4a + 3b + 2 =\n(6ab + 4a) + (3b + 2)\nthen factor out the common factor 2a in (6ab + 4a)\n(Note: to factor out is the same as dividing each term with the common factor)\n\nwe get:\n2a(3b + 2) + (3b + 2)\nNotice that in each term of the above we have a common factor of (3b + 2)\nBy dividing each term by (3b + 2) or factoring out (3b + 2) we have:\n(3b + 2)(2a + 1)