Question

Five distinct 2digit no are in a geometric progression. Find the sum of second and fourth digit

Answer 1

Given :  Five distinct 2digit no are in a geometric progression.  

To find : sum of second and fourth digit

Solution:

Let say  

Five Digits are

a ,  ar , ar²  , ar³  , ar⁴

ar⁴   < 100

a ≥ 10  

=> r⁴   <  10

=> r < 1.78

r ≠ 1

So r will be a rational number  Let say (p/q)   where p & q are co prime  & integers

now a (p/q)⁴ is an integer  

=> a must be  = kq⁴      k is an integer  ( as p & q are co prime)so q⁴  in denominator must be cancelled by a

Hence a can be  k(2)⁴   or k (3)⁴   because 4⁴ > 99  so   q can be 2 or 3 only

Hence a can be  16k   or   81k    

Case  1 :

q = 2

=> a = 16k  then  a (p/q)⁴   = kp⁴   then p can be  3 only  & k has to be 1

k > 1 will make  ar⁴ > 99

so r = p/q  = 3/2

a = 16      as k = 1

16  ,  24   ,  36  , 54  , 81  are the number

if q = 3

then a = 81k  hence k must be 1  =>  a = 81

a (p/q)⁴  =  p⁴     Then p must be 2    as  q = 3

so r = 2/3

Numbers

are 81  , 54  , 36 , 24  , 16

Sum of 2nd & fourth Digit in both cases = 24 + 54 = 78

78 is the sum of second and fourth digit

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