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

Hence a can be  k(2)⁴   or k (3)⁴  

Hence a can be  16k   or   81k    

q = 2

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

so r = p/q  = 3/2

a = 16      

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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Answer 2

Step-by-step explanation:

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

• Consider the terms in a G.P are a,ar,ar^2,ar^3,ar^4
• Since all are two digit numbers we can take
• a > 10
• and ar^4 < 99
• So r^4 < 99 / 10
• Or r < = 1.7
• Since nearest number is 1.5 (3/2) for 1.7 we get
• So m1 = a = 2^4 = 16
• So m2 = a r
•           = 16 x 3/2
•          = 24
• So m3 = a.r^2
•           = 16 x (3/2)^2
•           = 36
• So m4 = a r^3
•           = 16 x (3/2)^3
•            = 54
• So m5 = a r^4 = 16 x (3/2)^4
•                        = 81
• So the numbers are 16,24,36,54,81
• Now sum of second and fourth digit will be 24 + 54 = 78

Reference link will be

https://brainly.in/question/18708636