Question

Can someone help me with this question

Answer 1

if any fault please rectify..

Answer 2

Answer:

a₁₀= -150

-150=a+(10-1)d

-150=a+9d_______eq.1

s₂₀=n÷2(2a+(n-1)d)

-550=20÷2(2a+19d)

-550=10(2a+19d)

-550-10= 2a+19d

-560=2a+19d_______eq.2

using eq.1 and 2

-560=2a+19d×1

-150=a+9d      ×2

-560=2a +19d

-300=2a +18d

+        -      -  

-260=d

d= -260

put d= -260 in eq.2

-560=2a +19d

-560=2a+19(-260)

-560=2a-4940

-560+4940=2a

4380=2a

4380÷2=a

2190=a

a=2190

a₂=2190+(2-1)-260

a₂=2190-260

a₂=1930

a₃=2190+2(-260)

a₃=2190-520

a₃=1670

AP is  2190,1930,1670_______