The pth term of an AP is 1÷7(2p-1).Find the sum of its n terms
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Hi ,
pth term = tp = ( 2p - 1 ) / 7
t1 = ( 2 × 1 - 1 ) / 7 = 1/7
t2 = ( 2 × 2 - 1 ) / 7 = 3/7
t3 = ( 2 × 3 - 1 ) / 7 = 5/7
first term = a = t1 = 1/7
common difference = t2 - t1
d = 3/7 - 1/7
d = 2/7
sum of n terms = n/2 [ 2a + ( n - 1 ) d ]
Sn = n/2 [ 2 × ( 1/7 ) + ( n - 1 ) ( 2/7 ) ]
= ( n/2) ( 2/7 ) [ 1 + n - 1 ]
= ( n/7 )( n )
= n² / 7
Therefore ,
Sn = n²/7
I hope this helps you.
:)
pth term = tp = ( 2p - 1 ) / 7
t1 = ( 2 × 1 - 1 ) / 7 = 1/7
t2 = ( 2 × 2 - 1 ) / 7 = 3/7
t3 = ( 2 × 3 - 1 ) / 7 = 5/7
first term = a = t1 = 1/7
common difference = t2 - t1
d = 3/7 - 1/7
d = 2/7
sum of n terms = n/2 [ 2a + ( n - 1 ) d ]
Sn = n/2 [ 2 × ( 1/7 ) + ( n - 1 ) ( 2/7 ) ]
= ( n/2) ( 2/7 ) [ 1 + n - 1 ]
= ( n/7 )( n )
= n² / 7
Therefore ,
Sn = n²/7
I hope this helps you.
:)
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