can you please solve this problem from progressions
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Hey frnd , hope this helps!
a1 = 4/n+1 d = 8/n+1 - 4 /n+1 = 4/n+1
Substitute these values in the given formula
Sn = n/2 ( 2a1 + ( n-1) d)
Sn = n/2 [2 × 4/n+1 + (n-1) 4/n+1]
Sn = n/2 × 4 /n+1 [ 2 + n - 1 ]............by taking 4/n+1 common.
Sn = 4n / 2(n+1) [ n+1].................n+1 gets cancelled out .
Sn = 2n
So the sum of the n terms of the given AP is 2n.
a1 = 4/n+1 d = 8/n+1 - 4 /n+1 = 4/n+1
Substitute these values in the given formula
Sn = n/2 ( 2a1 + ( n-1) d)
Sn = n/2 [2 × 4/n+1 + (n-1) 4/n+1]
Sn = n/2 × 4 /n+1 [ 2 + n - 1 ]............by taking 4/n+1 common.
Sn = 4n / 2(n+1) [ n+1].................n+1 gets cancelled out .
Sn = 2n
So the sum of the n terms of the given AP is 2n.
Penmighter:
Welcome!
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