Latest Infosys Aptitude Question SOLUTION: 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.

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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

Asked In Infosys SUPREETHA S (5 years ago)

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Pradeep Sharma  (5 years ago)

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.

Is this solution Helpfull? Yes (3) | No (11)  

Barkha bhatia  (5 years ago)

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.

Is this solution Helpfull? Yes (3) | No (1)  

Aarthi  (5 years ago)

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,...

Is this solution Helpfull? Yes (2) | No   

sharad  (5 years ago)

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Is this solution Helpfull? Yes | No (8)  

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