Question

There are four numbers such that first three of them form an AP and the last three form a GP .The sum of the first and third number is 2 and that of second and fourth is 26. What are these number

Answer 1

Let the 1st 3 terms be a-d,a,a+d
Ac. to Q. a-d+ a+d=2 
Therefore, a=1
Now, let the last 3 terms be a,ar,ar2
​Ac. to Q. a+ar2= 26
a(1+r2)=26
r2 =25
Therefore, r= +5 and -5
Therefore, the no.'s are -3,1,5,25 or 7,1,-5,25

Answer 2

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}