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
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
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) ) = ...
Tanks alot
tnx man
Thanks 😂 🤣
Sn=n/2[2a+(n-1)d]
S18=549
S18=18/2[2a+(18-1)3]
549=9[2a+17(3)]
549=18a+459
549-459=18a
90/18=18a/18
a=5
Tn=a+(n-1)d
T56=5+(56-1)3
T56=5+165
T56=170
S32=n/2[2a+(n-1)d]
S32=32/2(10+31(3)]
S32=16(103)
S32=1648