# Sum of "n" terms of an Arithmetic Progression:

#### Sum of n terms of an Arithmetic Progression:

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 $$S = \frac{n}{2} \ \{a + a + (n - 1) \ d\}$$ $$\bbox[5px,border:1px solid black] { S = \frac{n}{2} \ \{2a + (n - 1) \ d\} }$$

#### If sum of n terms of a series is given then the $$n^{th}$$ term:

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.