Let the first three numbers be a₁ , a₁+d and a₁+2d
Let the last three numbers be a, ar and ar²
As there are total three numbers, thus we can say...
1st number = a₁
2nd number = a₁+d or a
3rd number = a₁+2d or ar
4th number = ar²
Thus we can say a₁+d = a ...eq(i)
According to question, 1st# + 3rd# = 2
or, a₁ + a₁ + 2d = 2
or, 2(a₁ + d) = 2
or, a₁ + d = 1
or, a = 1 [From eq(i)] ...eq(ii)
According to question, 2nd# + 4th# = 26
or, a + ar² = 26
or, a(1 + r²) = 26
or, 1 + r² = 26 [As a = 1, from eq(ii)]
or, r = ±5 ...eq(iii)
Thus 2nd# = a = 1
3rd# = ar = 5 or -5
4th# = ar² = 25
Thus by comparing 2nd# and 3rd# we can get d
d = a₃ - a₂
d = 5 - 1 or -5 - 1
d = 4 or -6
Therefore we know a₁ = a₂ - d
a₁ = 1 - 4 or 1 - (-6)
a₁ = -3 or 7
Thus there are two sets of numbers....
{-3, 1, 5, 25} and {7, 1, -5, 25}