the sum of three numbers in GP is 56 if we subtract 1, 7, 21 from these numbers in - Maths - Sequences and Series

Question

the sum of three numbers in GP is 56 if we subtract 1, 7, 21
from these numbers in that order, we obtain an AP find the
numbers

Santosh Kumar · Accepted Answer

Let the three numbers in G P be $\frac{a}{r}, a, a r$ Now, $\frac{a}{r} + a + a r = 56 \Rightarrow a (\frac{1}{r} + 1 + r) = 56 \Rightarrow a (\frac{1}{r} + r) = 56 - a$ Also, $\frac{a}{r} - 1, a - 7, a r - 21 are in A P$ As in a A P, common difference of the successive terms are same $So, (a - 7) - (\frac{a}{r} - 1) = (a r - 21) - (a - 7) \Rightarrow α - 7 - \frac{α}{r} + 1 = α r - 21 - α + 7 \Rightarrow 2 α - α (\frac{1}{r} + r) + 8 = 0 \Rightarrow 2 α - 56 + α + 8 = 0 [from equation (1)] \Rightarrow α = 16$ Now, substituting Î± = 16 in equation (1), $r = \frac{1}{2} or 2$ Hence, three numbers in G P = 32, 16, 8 or 8, 16, 32

the sum of three numbers in GP is 56. if we subtract 1, 7, 21 from these numbers in that order, we obtain an AP. find the numbers.