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

Answer 1

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

Answer 2

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