My Math Forum A.m a.p g.m g.p

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 May 5th, 2017, 11:55 PM #1 Newbie   Joined: May 2017 From: Pakistan Posts: 7 Thanks: 0 A.m a.p g.m g.p Please solve these 1) Insert n A.Ms between 1 and n² 2) b+c-3a/a , c+a-3b/b , a+b-c/c are in A.P prove that 1/a , 1/b , 1/c are also in A.P 3) if the roots of equation a(b-c)x² + b(c-a)x + c(a-b) = 0 are equal prove that a, b , c are in H.P 4) The sums of the first n terms of two A.Ps are in the ratio 3n+31 : 5n-3 find the ratio of their nth terms 5) if G1 and G2 are two G.Ms between b,c and A is their A.M then show that G1^3 + G2^3 = 2Abc 6) If a,b,c in G.P and x,y are the A.Ms of a, b and b, c respectively then prove that a/x + c/y = 2 7) An A.P has first term of 6 and fifth term of 18 find the sum of the n terms of the progression is greater than 2000 Find a G.M between 3 and 12 9) Sum to n terms the series 2+22+222....
 May 20th, 2017, 03:59 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,194 Thanks: 871 I am going to guess, because you did not say, that "A.M" is "arithmetic mean", "G.P." is "geometric proportion", and "G.M" is "geometric mean. Do you know the definitions of those? In the second exercise you have "b+c-3a/a". That means b+ c- (3a/a) which is the same as b+ c- 3. Did you mean (b+ c- 3a)/a? Please use parentheses! You have listed 9 arithmetic problems with no attempt shown to solve them yourself. If you have no idea how to even begin then you need to talk to your teacher. You have far worse problems than we can help you with.