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# The first and last terms of an AP are a and l respectively. If S be the sum of all the terms of the AP, the common difference is $\dfrac{{{l}^{2}}-{{a}^{2}}}{2S-\left( l+a \right)}.$(a) True(b) False

Last updated date: 17th Sep 2024
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Hint: We are given the first and the last term as a and l respectively and also the sum of n terms using the sum formula of an AP which is given as ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right].$ We first have to find the number of terms, i.e. (n) and then we will use the formula ${{a}_{n}}=a+\left( n-1 \right)d$ to find the common difference (d).

Complete step-by-step solution:
We are given that the first term of the AP is denoted by a and the last term is denoted by l. The sum of all the terms is represented by S. We know that the sum of all the terms (n terms) is given by the formula defined as ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right].$ We have ${{S}_{n}}$ as S.
So using the above formula, we will find the value of n. Putting S in place of ${{S}_{n}}$ we will get,
$S=\dfrac{n}{2}\left[ a+l \right]$
Now, we will solve for n.
$2S=n\left( a+l \right)$
$\Rightarrow n=\dfrac{2S}{a+l}$
So, we got the value of n as $\dfrac{2S}{a+l}.$
Thus we have total $n=\dfrac{2S}{a+l}$ terms in AP.
Now, we have the last term given to as l and we know that the last term with the total terms as n is defined by the form $l=a+\left( n-1 \right)d.$
Now, we will place $n=\dfrac{2S}{a+l}$ in the above equations, then we will get,
$\Rightarrow l=a+\left( \dfrac{2S}{a+l}-1 \right)d$
Simplifying the above terms,
$\Rightarrow l=a+\left( \dfrac{2S-\left( a+l \right)}{a+l} \right)d$
Now, we will solve for d, so we will get,
$\Rightarrow l-a=\left( \dfrac{2S-\left( a+l \right)}{a+l} \right)d$
Taking a + l to the other side, we will get,
$\Rightarrow \left( l-a \right)\left( a+l \right)=\left[ 2S-\left( a+l \right) \right]d$
We can write, $\left( l-a \right)\left( a+l \right)={{l}^{2}}-{{a}^{2}}$
So, we get,
${{l}^{2}}-{{a}^{2}}=\left[ 2S-\left( a+l \right) \right]d$
Divide both the sides by $\left[ 2S-\left( a+l \right) \right],$ we have,
$\Rightarrow \dfrac{{{l}^{2}}-{{a}^{2}}}{2S-\left( a+l \right)}=d$
So, we get the common difference as $\dfrac{{{l}^{2}}-{{a}^{2}}}{2S-\left( a+l \right)}.$
Hence, we were given the correct common difference.
Therefore, option (a) is the right option.

Note: While calculating, students need to be very careful as $\dfrac{2S}{a+l}-1\ne \dfrac{2S-1}{a+l},$ we have to take the LCM to subtract. There is no need to get confused over the fact of a + l or l + a, as both the terms are the same $\left( l-a \right)\left( a+l \right)=\left( l-a \right)\left( l+a \right)$ which gives us ${{l}^{2}}-{{a}^{2}}.$