The sum of 18 term of an ap is 15.549 given that the common difference is 3 find the 56th term and also the sum of its 32term
Sum of n terms in AP
Sn = ( n / 2 ) [ 2 a1+ ( n - 1) d ]
where
a1 = the initial term
d = the common difference of successive members
In this case:
n = 18
d = 3
S18 = ( 18 / 2 ) ∙ ( 2 a1+ 17 d ) = 15549
9 ∙ ( 2 a1+ 17 ∙ 3 ) = 15549
9 ∙ ( 2 a1+ 51 ) = 15549
18 a1 + 459 = 15549
Subtract 459 to both sides
18 a1 = 15090
a1 = 15090 / 18 = 6 ∙ 2515 / 6 ∙ 3
a1 = 2515 / 3
n-th term in AP
an = a1 + ( n - 1 ) d
a56 = a1 + 55 d
a56 = 2515 / 3 + 55 ∙ 3 = 2515 / 3 + 165 = 2515 / 3 + 495 / 3
a56 = 3010 / 3
Sn = ( n / 2 ) [ 2 a1+ ( n - 1) d ]
S32 = ( 32 / 2 ) [ 2 ∙ 2515 / 3 + 31 ∙ 3 ]
S32 = 16 ∙ [ 2 ∙ 2515 / 3 + 31 ∙ 3 ] = 16 ∙ ( 5030 / 3 + 93 ) =
16 ∙ ( 5030 / 3 + 279 / 3 ) = 16 ∙ 5309 / 3
S32 = 84944 / 3
Sn=n/2[2a+(n-1)d]
15.549= 18/2[2a+(18-1)3
15.549= 9[2a+(17×3)]
15.549= 9[2a+51]
15.549= 18a+459
18a= 15.549-459
18a= -443.451
a= -443.451/18
a= -24.636
nth term=a+(n-1)d
56th term= -24.636+(56-1)3
=-24.636+(55×3)
=-24.636+165
=140.364
S32= 32/2[2×-24.636+(32-1)3
=16[-49.272+(31×3)
=16[-49.272+93]
=16×43.728
=699.648
thus, the 56th term is 140.364 and the sum of the first 32 terms is 699.648