Question

The sums of three numbers in G.P is 14. If the first two terms are each increased by 1 and the third term decreased by 1,the resulting numbers are in A.P.Find the nubers

Answer 1

$\huge{Hey Mate!!!}$

☆☞ Here is ur answer ☜☆

☆☞ let the required no.s be a, ar, ar^2...
actq,
a + ar + ar^2 = 14
a (1 + r + r^2 )= 14
a = 14 / (1+ r + r^2 ).....................(i)

and, a+1, ar+1, ar^2-1 are in ap
therefore by using 2b=a+c we'll have,
2(ar+1) = a+1+ar^2-1
2ar+2 = a+ar^2
a-2ar+ar^2 = 2
a(1-2r+r^2) = 2
now from eq(i), we'll replace a by 14 / (1+r+r^2)
on solving we'll have,
r=2,1/2 and a=2,8

therefore required numbers will be (2,4,8) OR (8,4,2).......


HOPE IT HELPS!!!

Answer 2

2(ar+1) = a+1+ar^2-1

2ar+2 = a+ar^2

a-2ar+ar^2 = 2

a(1-2r+r^2) = 2

14 / (1+r+r^2)

r=2,1/2 and a=2,8