Question

in an arithmetic sequence, 5t5 = 9t, then prove that t14 = 0

Answer 1

Answer:

Given sequence is,

3,6,12,24,...…

first term of this A.P is  a1=3

second term of this A.P is  a2=6

third term of this A.P is  a3=12

the condition for an sequence to be an A.P is their must be a common difference (i.e.,d=an+1−an)

putting n=1 in above equation

d=a2−a1=6−3=3

putting n=2 in above equation

d=a3−a2=12−6=6

as we can see we get two values of d, 

but in an A.P their must be a single value of d which means for this sequence  we can't define a common difference.

hence this sequence  not form an A.P