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There are four numbers such that the first three of them form an arithmetic

There are four numbers such that the first three of them form an arithmetic sequence and the last three form a geometric sequence. The sum of the first and the third term is 2 and that of second and fourth is 26. Find the numbers

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A)
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}
 
 
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