Let a, ar, $ar^2$ be three numbers in G.P. whose sum is 56, then a + ar + $ar^2$ = 56 ⇒ a(1 + r + $r^2$) = 56 .....(i) We are given, if 1, 7, 21 are subtracted from a, ar, $ar^2$ respectively, then they form an A.P. i.e., a – 1, ar – 7, $ar^2$ – 21, ...... form an A.P. Then 2(ar – 7) = ($ar^2$ – 21) + (a – 1) ⇒ 2ar – 14 = $ar^2$ + a – 22 ⇒ $ar^2$ + a – 2ar = – 14 + 22 ⇒ a($r^2$ – 2r + 1) = 8 .... (ii) Dividing (i) by (ii), we get $rac{a(1+r+r^2)}{a(r^2-2r+1)}$ = $rac{56}{8}$ ⇒ $rac{1+r+r^2}{1-2r+r^2}$ = 7 ⇒ 1 + r + $r^2$ = 7 - 14r + $7r^2$ ⇒ $6r^2$ - 15r + 6 = 0 ⇒ $2r^2$ - 5r + 2 = 0 ⇒ (2r - 1) (r - 2) = 0 ⇒ r = $rac{1}{2}$ , 2 If r = $rac{1}{2}$ , then a (1 + r + $r^2$) = 56 ⇒ $a\left(1+\frac{1}{2}+\frac{1}{4}\right)$ = 56 ⇒ $a\left(\frac{4+2+1}{4}\right)$ = 56 ⇒ a = $rac{56 imes 4}{7}$ = 32 Then, the numbers are 32, 16, 8 If r = 2, then a(1 + r + $r^2$) = 56 ⇒ a(1 + 2 + 4) = 56 ⇒ a = $rac{56}{7}$ = 8 Then, the numbers are 8, 16, 32. Hence, required numbers are 8, 16, 32.