Write thenth term of an AP, the sum of whose n term is sn
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we know, difference between two consecutive terms of an arithmetic progression (AP) must be same.
e.g., if a is the first term and d is the common difference of an AP.
then, series is : a , a + d , a + 2d , a + 3d , a + 4d...... are in AP
now, nth term = T_nT
n
= a + (n - 1)d
sum of n terms = S_nS
n
= a + (a + d) + (a + 2d) + ..... + [a + (n - 1)d]
= (a + a + a ....n times ) + [d + 2d + 3d + .... + (n - 1)d ]
= na + d[1 + 2 + 3 + ..... + (n -1)]
= na + d[(n - 1)(n)/2 ]
= n/2 [ 2a + (n - 1)d]
hence, S_nS
n
= n/2 [2a + (n - 1)d ]
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