Arithemetic progression
posted by Anonymous .
the sum of 18term of an A.P is549 given the common difference is 3 find the 56th term and the sum of its 32term

term(18) = a+ 17d
but d = 3 and term(18) = 549
a + 17(3) = 549
a = 498
term(56) = a + 55d = .....
sum(32) = (32/2)(2(498) + 31(2) ) = ... 
unless he meant that S18 = 549
In that case,
18/2(2a+17*3) = 549
a = 5
T56 = 5+55*3 = 170
S32 = 32/2(2*5+31*3) = 1648