Question

Can you solve this?Hard One

Answer 1

Let 1st 3 numbers be a-d, a, a+d as it's in AP.
where d=common difference
It is given that sum of 1st and 3rd term=2
a-d+a+d=2
a=1
Again, it is given that the last 3 numbers are in G.p
so, let the last 3 terms be a, ar, ar²
And it's given that sum of 2nd and 4th terms is equal to 26.
=>a+ar²=26
=>r²=25
=>r=±5
If r=5, then the last 3 terms will be, a,ar, ar²=1, 5,25 and here d will be 3rd term-2nd term=5-1=4, so, 1st term will be, a-d=1-4=-3
so, the sequence is -3,1,5,25
If r=-5,3rd term=-5, 4th term=25, 2nd term=1 d=-5-1=-6, 1st term=a-d=1-(-6)=7
then, the sequence will be, 7,1,-5,25
There will be 2 sequence, (-3, 1, 5, 25) &(7,1,-5,25)