in an AP,the sum of first ten terms is -150 and the sum of its next ten terms is -550.Find the AP.

S_n= n/2 [2a + (n-1)d]

S₁₀ = 5 [2a + 9d]                                  

-150 ³⁰=  5  ¹[2a + 9d]

30 = 2a + 9d

2a = 30 - 9d            _______________(i)

S₂₀ = 10[2a + 19d]

S₂₀ =  10[2a + 19d]

-550 - 150 = 10 [30 - 9d + 19d]                               [from (i)]

-700 / 10 = 30 + 10d 

-70 - 30 = 10d

-100 = 10d

d = -10

putting value of d in (i)

2a = 30 - 9(-10)

2a = 30+ 90

a = 60

AP = 60 , 50 , 40 ,......................

 S_n= n/2 [2a + (n-1)d]

S₁₀ = 5 [2a + 9d]      

-150=5[2a+9d]

-30=2a+9d ---(1)

according to second condition for the next 10 term 

for this the 11 th term of the ap will the first term

i.e a+10d

so by the formula 

S_n= n/2 [2a + (n-1)d]

put n=10,a-->a+10d 

-550=5[2(a+10d)+9d]

-110=2a+20d+9d

-110=2a+29d

put the value of 2a from (1) equation i.e 2a=-30-9d

-110=-30-9d+29d

-110+30=20d

-80=20d

d= -4

2a=30-9*(-4)

2a=30+36

2a=66

a=18

ap is 18,14,10

 S_n= n/2 [2a + (n-1)d]

S₁₀ = 5 [2a + 9d]      

-150=5[2a+9d]

-30=2a+9d ---(1)

according to second condition for the next 10 term 

for this the 11 th term of the ap will the first term

i.e a+10d

so by the formula 

S_n= n/2 [2a + (n-1)d]

put n=10,a-->a+10d 

-550=5[2(a+10d)+9d]

-110=2a+20d+9d

-110=2a+29d

put the value of 2a from (1) equation i.e 2a=-30-9d

-110=-30-9d+29d

-110+30=20d

-80=20d

d= -4

2a=30-9*(-4)

2a=30+36

2a=66

a=33

ap is 33,29