Question

the first and last term of AP are a and l respectively. If S is the sum of all the terms of the AP and the common difference is given by (l^2-a^2)/k-(l+a) , then k = ?

Answer 1

The first and last term of AP are a and l respectively. If S is the sum of all the terms of the AP and the common difference is given by $(1^2 - a^2)/k - (l + a)$, then $k = ?$

Solution

Let the common difference and number of terms of AP be $d$ and $n$ respectively.

Last term of $\mathrm{AP} = n^{th}$ of the AP, $a_{n}$

$$\begin{array}{l} \rightarrow a _ {n} = / (G i v e n) \\ \rightarrow a + (n - 1) d = l \rightarrow n = \frac {l - a}{d} + 1 \quad \dots (1) \end{array}$$

Sum of $n$ terms of $A.P. = S(Given)$

$$\begin{array}{l} \frac {n}{2} [ 2 a + (n - 1) d ] = S \\ \rightarrow \frac {1}{2} \left(\frac {l - a}{d} + 1\right) \left[ 2 a + \left(\frac {l - a}{d} + 1 - 1\right) d \right] = S \quad \\ \rightarrow \frac {1}{2} \left(\frac {l - a}{d} + 1\right) [ 2 a + (l - a) ] = S \\ \rightarrow \frac {1}{2} \left(\frac {l - a}{d} + 1\right) (a + l) = S \\ \rightarrow \frac {l - a}{d} + 1 = \frac {2 S}{l + a} \\ \rightarrow \frac {l - a}{d} = \frac {2 S}{l + a} - 1 = \frac {2 S - (l + a)}{l + a} \\ \rightarrow d = \frac {(l + a) (l - a)}{2 S - (l + a)} \end{array}$$

Comparing with $d = \frac{l^2 - a^2}{k - (l + a)}$, we get $k = 2S$

Thus, the value of $k$ is 2S.

Answer: the value of $k$ is 2S.