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
