In an AP the sum of first ten terms is -150 and the sum of its next 10 terms is -550. Find the AP.
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This is the method to find first term and common difference. Now AP can be found by a, a+d, a+2d,a+3d,.........
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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.
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.
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