Let's consider the first term of an AP is "a", common difference is "d", \(n^{th}\) term is "l" and the sum of n terms is "S" then,
$$ S = a + (a + d) + (a + 2d) + \\.....+ (l - d) + l.....(1) $$
because (n - 1) term will be less than "d" from the last term "l". Let's write the above series in reverse sequence.
$$ S = l + (l - d) + (l - 2d) + \\.....+ (a + d) + a.....(2) $$
By adding equations (1) and (2), we get
$$ 2S = (a + l) + (a + l)... \\.....(a + l) + (a + l) $$
Here the number of terms are "n", Hence $$ 2S = n \ (a + l) $$ $$ S = \frac{n}{2} \ (a + l).....(3) $$ as we know \(l = a + (n - 1) \ d......(4)\)
By putting the value of "l" from equation (4) to equation (3), we get $$ S = \frac{n}{2} \ \{a + a + (n - 1) \ d\} $$ $$ \bbox[5px,border:1px solid black] { S = \frac{n}{2} \ \{2a + (n - 1) \ d\} } $$
Let sum of n terms of a series is \(T_n\)
Then sum the (n - 1) terms of a series will be = \(T_{n - 1}\)
Hence, \(n^{th}\) term (l) of a series $$ \bbox[5px,border:1px solid black] { l = T_n - T_{n - 1} } $$
Example(1): Find the sum of a given series? $$ 1 + 4 + 7 + .....+ 16 $$
Solution: from the given series, a = 1, d = 4 - 1 = 3, n = 16
Let the sum of the series is "S" then, $$ S = \frac{n}{2} \ {2a + (n - 1) \ d} $$ $$ = \frac{16}{2} \ {2 \times 1 + (16 - 1) \times 3} $$ $$ = 8 \ {2 + 45} $$ $$ = 8 \times 47 = 376 $$ Hence the sum of the series (S) is 376
Example(2): If the sum of n terms of an Arithmetic series is 10n. Then find the \(n^{th}\) term of that series?
Solution: Given, sum of n terms of an arithmetic series $$ T_n = 10 \ n $$ Hence sum of (n - 1) term of that series $$ T_{n - 1} = 10 \ (n - 1) $$ So the \(n^{th}\) term of that series $$ l = T_n - T_{n - 1} $$ $$ l = 10 \ n - 10 \ (n - 1) $$ $$ = 10 \ n - 10 \ n + 10 $$ $$ l = 10 $$ Hence \(n^{th}\) term of that series is 10.