Given an A.P. whose terms are all positive integers. The sum of its first nine terms is greater then 200 and less than 220. If the second term in it is 12, then its 4th term is. :

a+d=12

sum of n terms in an A.P = N/2[2a+(n-1)d] So, sum to 9 terms= 9/2[2a+8d] According to given question, 200<9/2[2a+8d]<220 = 200<9[a+4d]<220 = 200/9

sum of n terms in an A.P = N/2[2a+(n-1)d] So, sum to 9 terms= 9/2[2a+8d] According to given question, 200<9/2[2a+8d]<220 = 200<9[a+4d]<220 = 200/9