Find the fifth term of the arithmetic sequence whose third term is 29 and whose twenty-ninth term is 289.
enter the numerical answer in the box below - no additional words or punctuation is needed. (example: 25)

Let first term is a and common difference is d
third term=29
so a+2d=29 eq.1
twenty ninth term=289
a+28d=289 eq.2
solving eq.1 and 2
we get 26d=260
d=10
So from eq.1
putting the value of d
a+2(10)=29
a+20=29
a=9
So fifth term is a+4d
=9+4(10)
=9+40
=49

Ithink the right answer could be 50%