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# 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. :

7 years ago
a+d=12
sum of first 9 terms 200<a+8d<220
200-12<a+8d-a-d<220-12
188<7d<208
27<d <29
taking d=28, 4th term is a+3d=-16+3*28 ,

b.tech, iit delhi

sai krishna
20 Points
7 years ago
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
sai krishna
20 Points
7 years ago
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