A.P.

[eco201611]

The First term of an arithmetic progression is a and common dierence
is d  (0, 1). Suppose the k-th term of this arithmetic progression
equals the sum of the infite geometric progression whose first term is
a and common ratio is d. If a > 2 is a prime number, then which of
the following is a possible value of d?

[Dr. Strange]

You have wrote "which of the following" but have not provided with the options.Provide options and I can try.

[eco201611]

options are
a)1/2
b)1/3
c)1/5
d)1/9

[Dr. Strange]

So,
a+(k-1)d= a/(1-d)

solving we have, a=(k-1)* (1-d)

now we look at options:
for d=1/3           ->         (1-d)=2/3
since (k-1) and ' a '  both are integers  so a=(k-1)*2/3
so k-1 should be divisible by 3 and the product is multiplied by 2 to get ' a '
so 'a' is a multiple of 2 and since it is given that 'a' is prime and greater than 2 so its not possible

similar is the case with d=1/5 or d=1/9

only option left is d=1/2
1-d=1/2
so a=(k-1)* 1/2
if k=7 we get a=3 which is prime
or if k=15 ,a=7 which is prime


so we can get infinite prime a's for d=1/2
and no prime a>2 for other options

so answer is d=1/2  

[eco201611]

thanks a lot