The sum of first ten terms of an A.P is 155 and the sum of first two terms of a G.P is 9. The first term of the A.P is equal to the common ratio of G.P and the first term of the G.P is equal to common difference of the A.P which can be the A.P as per the given conditions

• The sum of 10 terms of an A.P can be represented as Sn=155
  After evaluation, we can get the equation
  31=2r+9d-------(1)
  The sum of two terms of G.P can be represented as Sn=9
  After evaluation, we can get the equation =
  9=a(r+1)----(2)
  Given that,
  a of A.P= r of G.P
  a of G.P = d of A.P
  So we substitute this in equation (1) and (2)
  31=2r+9d------(3)
  9=d(r+1)----(4)
  Now after manipulation, we can get
  (31-9d)/2=r-----(5)
  Now substituting the value of equation (5) in equation (4) we get the values of d
  d=3, d=2/3
  So we take d=3.
• 6 years agoHelpfull: Yes(3) No(11)
• Let the ap be a, a+d,a+2d.............a+9d and the gp be a1, a1r.....
  sum of ap(155) =10(a+a+9d)/2
  so, 2a+9d=31
  
  given, a1+a1r=9 , a=r and a1=d
  a1(1+r)=9
  substituting values:-
  d(1+r)=9 and 31=2r+9d
  on solving these two equations we get d=3
  
  so ans is d=3.
• 6 years agoHelpfull: Yes(3) No(1)
• Let the A.P be a,a+d,a+2d,...
  and G.P be b,br,br^2,..
  Given ,
  Sum of 10 terms of A.P=155
  so,n/2(2a+(n-1)d)=155
  10/2(2a+9d)=155
  2a+9d=31---(1)
  b+br=9
  substitute b=d,r=a
  d+ad=9----(2)
  by sloving 1 and 2
  we get,a=2,25/2 and d=3,2/3
  so A.P can be 2,5,8,11,.. (or)
  25/2,79/6,83/6,...
• 6 years agoHelpfull: Yes(2) No(0)
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• 6 years agoHelpfull: Yes(0) No(8)