Question

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

Answer 1

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) ) = ...

Answer 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

Answer 3

tnx man

Answer 4

Tanks alot

Answer 5

Thanks 😂 🤣

Answer 6

Thanks again for your help

Answer 7

thanks a lot I am gratefull

Answer 8

Thanks

Answer 9

Because of the answer is not working

Answer 10

Thanks a

lot

Answer 11

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

Answer 12

Thanks alot