Radha says" 1, 1, 1, ..... are in A.P. and also in G.P.". Do you agree with Radha?
Give reason
Answers
Answer:
Yes, I agree. As a reason :
Given sequence is 1 , 1 , 1 .... , in this situation we can easily get that the common difference between the terms is 0, due to which, all terms of the AP will be 1, since 1 + 0 = 1 + 2( 0 ) = 1 + 3( 0 )...= 1.
Therefore the sequence 1 , 1 , 1 ...... is in arithmetic progression ( A.P ).
An another reason :
From the properties of arithmetic progressions we know that the twice of middle is equal to the sum of remaining terms.
So, if 1 , 1 , 1 ..... are in AP.
= > twice of middle term = sum of remaining terms
= > 2( 1 ) = 1 + 1
= > 2 = 2
Right hand side is equal to left hand side.
Thus the given sequence 1 , 1 , 1.... is in AP.
Proof that the sequence is in GP :
Given sequence is 1 , 1 , 1 .... , in this situation we can easily get that the common ratio between the terms is 1, due to which, all terms of the AP will be 1, since 1( 1 )= 1( 1 )² = 1( 1 )³...= 1.
Hence, this is in GP too.
Another reason :
From the properties of Geometric progressions, square of middle term is equal to product of other terms and here square of 1 is 1 and the product of 1 & 1 is also 1.
Hence they are in GP too.
Thus,
Radha is correct. I agree with her.