Question

find the sum of first 32 term of AP 8,3,-2​ . . .

Answer 1

Answer:

given that ap ={8,3,-2....}

Step-by-step explanation:

we know that SN=n/2(2a+(n-1)D)

n=32

a=8

d=3-8=-5

1. s32=32/2(2×8+(32-1)-5)

s32=16(16+(31)-5). [31×-5=-155]

s32=16(16+(-155))

s32=16(16-155). [155-16=139]

s32=16(-139)

[16×- 139=2124]

s32=2124

. . s32 = 2124

.

Answer 2

Answer:

Here a=8,d=3-8= -5

As we know,

Sum of nth term = n/2[2a+(n-1)d]

Therefore for 32 terms

Sum= 32/2[2×8+(32-1)(-5)]

Sum= 32/2[2×8+(32-1)(-5)] =16[16+31×(-5)]

Sum= 32/2[2×8+(32-1)(-5)] =16[16+31×(-5)] =16[16-105]

Sum= 32/2[2×8+(32-1)(-5)] =16[16+31×(-5)] =16[16-105] =16×(-89)

Sum= 32/2[2×8+(32-1)(-5)] =16[16+31×(-5)] =16[16-105] =16×(-89) = -1424