Question

How do you find $a_n$ for the arithmetic series ${{S}_{16}}=856$ and ${{a}_{1}}=76$?

Answer 1

Hint: This question can be solved by using the formula for the ${{n}^{th}}$ term of an AP which is given by the expression ${{a}_{n}}=a+\left( n-1 \right)d$, where a is the first term and d is the common difference. According to the question, the first term is given as ${{a}_{1}}=76$. For determining the value of d we have to use the sum of the $16$ terms of the AP which is given as ${{S}_{16}}=856$. For this, we have to use the formula for the sum of the n terms of an AP which is given by ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$. On substituting $n=16$, $a=76$ and ${{S}_{16}}=856$ into this formula, we will get the value of d. Finally, substituting the values of a and d into the expression ${{a}_{n}}=a+\left( n-1 \right)d$, we will get the final answer.

Complete step by step solution:
We know that the ${{n}^{th}}$ term of an AP is given by the formula
$\Rightarrow {{a}_{n}}=a+\left( n-1 \right)d........\left( i \right)$
Therefore, we need to determine the values of a and d.
According to the question, we have the first term as ${{a}_{1}}=76$. So we can write
$\Rightarrow a=76.........\left( ii \right)$
Also, the sum of the sixteen terms of the AP is given as
$\Rightarrow {{S}_{16}}=856.......\left( iii \right)$
We know that the formula for the sum of the n terms of an AP is given by
$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
On substituting $n=16$ we get
$\begin{align}
  & \Rightarrow {{S}_{16}}=\dfrac{16}{2}\left( 2a+\left( 16-1 \right)d \right) \\
 & \Rightarrow {{S}_{16}}=8\left( 2a+15d \right) \\
\end{align}$
Now, we substitute (ii) and (iii) into the above equation to get
\[\begin{align}
  & \Rightarrow 856=8\left( 2\left( 76 \right)+15d \right) \\
 & \Rightarrow 856=8\left( 152+15d \right) \\
 & \Rightarrow 856=1216+120d \\
\end{align}\]
Subtracting \[1216\] from both the sides, we get
$\begin{align}
  & \Rightarrow 856-1216=1216+120d-1216 \\
 & \Rightarrow -360=120d \\
\end{align}$
Now, on dividing $120$ both the sides, we get
\[\begin{align}
  & \Rightarrow \dfrac{-360}{120}=\dfrac{120d}{120} \\
 & \Rightarrow -3=d \\
 & \Rightarrow d=-3.......\left( iv \right) \\
\end{align}\]
Finally, on substituting (ii) and (iv) in (i) we get
\[\begin{align}
  & \Rightarrow {{a}_{n}}=76-3\left( n-1 \right) \\
 & \Rightarrow {{a}_{n}}=76-3n+3 \\
 & \Rightarrow {{a}_{n}}=-3n+79 \\
\end{align}\]
Hence, we have found the expression for ${{a}_{n}}$ as \[-3n+79\].

Note: We must remember the formulas for the ${{n}^{th}}$ term, the sum of n terms for solving the questions related to the arithmetic progression. We must not forget the factor of $\dfrac{n}{2}$ in the formula for the sum of the n terms of the AP which is given by ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$.