A.P in the given problem is 8,8+d,8+2d...
And it is given that the 1st,5th and 8th terms of the A.P are equal to the
,2nd and 3rd terms of a G.P.
First term of A.P is equal to the first term of G.P.
t5(fifth term of A.P)=a2(second term of G.P)
2r=2+d after dividing both sides by 4
This is the equation which is the relation between ratio and the common
Let this be eq1.
a3=t8 as per the given information
i.e ar.r=t+7d where a=t=8
as r=2+d/2 as per the eq1
we get 8r.2+d/2=8+7d
-3d=-4dr after cancelling -d on both sides we get
and finally r=3/4, so the common ratio is 3/4 in the G.P
By substituting this value in eq1 we get the value of d which is -1/2.
And by applying the formulae for sum to infinity of the G.P and sum of first 8terms of the A.P which are 16/3 and 9/2 respectively.