the first term of an ap is 3 and 11th term is 18 find the number of the term in the progression if the sum is 81
18 = 3 +(11-1)d
15 = 10 d
d = 1.5
so
Tn = 3 + (n-1) d
Sum n = (n/2) (2 a +(n-1)d ) which is the average of the first and the last
81 *2 = 162 = n ( 2*3 + 1.5 n - 1.5) = n (4.5 +1.5n) =1.5 n^2 + 4.5 n
1.5 n^2 + 4.5 n - 162 = 0
3 n^2 + 9 n - 324 = 0
try n = 9