Questions & Answers

Question

Answers

(a) True

(b) False

Answer

Verified

129.6k+ views

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.