Math, asked by rabiaboulhousn, 1 year ago

Determine the numbers a,b and c such that a,b,c are three consecutive terms of a geometric sequence and an arithmetic sequence and a×b×c=27

vishalkumar2806: Is the ques correct

rabiaboulhousn: No

vishalkumar2806: ok

rabiaboulhousn: sorry I mean yes the question is correct

Answers

Answered by vishalkumar2806

Cannot be determined..

complex solution are formed

vishalkumar2806: I cannot edit the app closes

rabiaboulhousn: ??

vishalkumar2806: Ok

vishalkumar2806: Is my ans correct

rabiaboulhousn: I don't know actually but I guess yes

rabiaboulhousn: can someone else help me

vishalkumar2806: Can you repost the ques

rabiaboulhousn: determime the numbers a,b,c such that a,b,c are three consecutive terms of a geometricsequence and an arithmetic sequence and abc=27

vishalkumar2806: Yha nhi....I Cannot edit my solution

rabiaboulhousn: oopsss

Answered by Anonymous

Answer:

Step-by-step explanation:

We can write the following statement, assuming the progression is n as -

a×(an)×(an×n) = 27 giving

(a*a*a)(n*n*n) = 27

The cube root of 27 is 3.

So either a or n must be 3 and the other number must be 1.

So either a=3 and n=1 or a=1 and n=3

Giving either: 1,3,9, total 13 or

3,3,3, total 9.

So the smallest possible sum is 9.

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