For a given A.P., for example 1,2,3,.... what is the formula to find the sum of the nth to kth elements ?
Like, if I want to find the sum from 5 to 10, what should I do ?
Should I use the standard formula,to find the sum till kth element and then deduct the sum till the nth element ?
If this is not the right approach, what is right ?
On
Almost. You sum all terms until the $k$th one, and then you subtract from that the sum until the $(n-1)$th one.
On
(I'm going to assume the last term in the sum is inclusive)
There are multiple ways to do this, the one you give is reasonable if you make sure that you subtract the sum up to the $(n - 1)$th element instead of the $n$th so that the $n$th term is still in the sum.
Example: $4 + 5 + 6 = (1 + 2 + 3 + 4 + 5 + 6) - (1 + 2 + 3)$.
You can also convert it to a smaller sum by rearranging the sum a bit:
Example: $4 + 5 + 6 = (3 + 3 + 3) + (1 + 2 + 3) = (4 - 1)(6 - 4 + 1) + (1 + 2 + 3)$
You have the right idea, just not the right formula or notation.
Suppose our arithmetic sequence is $$a_1, a_2, a_3, \ldots, a_n, \ldots, a_k, \ldots,$$ where $n \le k$, and let $d = a_2 - a_1$ be the common difference. Then we have for every positive integer $m$, $$a_m = a_1 + (m-1)d. \tag{1}$$ Now also define $$S_m = \sum_{i=1}^m a_i = \sum_{i=1}^m a_1 + (i-1)d = m a_1 + d \frac{(m-1)m}{2} = \frac{m}{2}(2a_1 + (m-1)d) = \frac{m(a_1 + a_m)}{2}. \tag{2}$$ This is the formula for the sum of the first $m$ terms. So the sum of the terms from $a_n$ to $a_k$, inclusive, is simply $$S_k - S_{n-1} = \frac{k(a_1 + a_k)}{2} - \frac{(n-1)(a_1 + a_{n-1})}{2}. \tag{3}$$ But this formula assumes you know the values of $a_1$, $a_{n-1}$, and $a_k$. What if you have only $n$, $k$, $a_n$, and $a_k$? For instance, if I give you $$n = 5, \quad k = 11, \quad a_n = a_5 = 38, \quad a_k = a_{11} = 80,$$ how do you proceed? Well, we can first find $d$ with the formula $$d = \frac{a_k - a_n}{k - n}, \tag{4}$$ which in our case yields $$d = \frac{80 - 38}{11 - 5} = \frac{42}{6} = 7.$$ Then we can find $a_1$ using Equation $(1)$ above: $$a_5 = 38 = a_1 + (5-1)7,$$ hence $a_1 = 10$. Now we can substitute everything in Equation $(3)$ to get $$S_k - S_{n-1} = \frac{11(10+80)}{2} - \frac{(5-1)(10+38-7)}{2} = 413.$$ And we can check this works by adding up all the terms manually: $$a_5 + a_6 + \cdots + a_{11} = 38 + 45 + 52 + 59 + 66 + 73 + 80.$$