Question

The sum of first 10 terms of an ap is - 150 and the sum of its next 10 terms is -550. Find the AP.

Answer 1

Let to sum of first 10 terms of an ap =Sn

the sum of its next 10 terms of an ap =sn'



S₁₀ = -150

S₁₀ + S'₁₀ = S₂₀

Given :-
S'₁₀ = -550

Therefore ,
S₂₀ = -150 -550
      = -700
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Lets form two linear equations and solve it :-

S₁₀ = -150
n = 10

Sn = n/2 [ 2a + (n-1)d]
-150= 10/2 [ 2a + 9d ]
-150 = 5 [ 2a + 9d ]
-30 = 2a + 9d

2a + 9d = -30          (Equation 1)

===========
S₂₀ = -700
numbers of terms n = 20

Sn = n/2 [ 2a + (n-1)d]
 -700   = 20/2 [ 2a + 19d ]
-700  = 10 [ 2a + 19d ]
-70 = 2a + 19d

2a + 19d = -70               (-equation 2)

solving equations 1 and 2 :-

(2) - (1) 

   2a + 19d = -70
  2a + 9d = -30
--------------------------------
10d = -40

d = -40/10 = -4

Substitute the value of d in equation;


2a + 9d = -30

2a -36 = -30

2a = -30+36

a = 6/2 = 3

a=3
---------------------------------------------

According to the question:-


first term (a )= 3
Common difference ( d)= -4

Arithmetic progressions:

3,-1,-5,-9



Thanks!!!

‎‎‎‎‎‎‎‎‎‎

Answer 2

Hey there !!

Let a be the first term and d be the common difference of the given AP .

S₁₀ = -150.

⇒ Sn = n/2 [ 2a + (n-1)d]

⇒ S₁₀ = 10/2 [ 2a + ( 10 - 1 ) d ].

⇒ -150= 10/2 [ 2a + 9d ]

⇒ -150 = 5 [ 2a + 9d ]

⇒ -30 = 2a + 9d

⇒2a + 9d = -30...........(1)

Clearly, the sum 20 term = - 150 + (-550) .

⇒ S₂₀ = -700

⇒ Sn = n/2 [ 2a + (n-1)d]

⇒ S₂₀ = 20/2 [ 2a + ( 20 - 1 )d ] .

⇒ -700   = 20/2 [ 2a + 19d ]

⇒ -700  = 10 [ 2a + 19d ]

⇒ -70 = 2a + 19d .

⇒ 2a + 19d = -70........(2)

Substracting 1 and 2 , we get 

  2a + 19d = -70

 2a + 9d = -30

-       -     +

____________

⇒ 10d = -40

⇒ d = -40/10 = -4

Put the value of d in equation 1.

2a + 9d = -30

⇒ 2a -36 = -30

⇒ 2a = -30+36

⇒ a = 6/2 = 3

a = 3

d = -4

Hence, AP is 3,-1,-5, - 9 ....

THANKS

#BeBrainly.