Question

How many terms are in the sequence $4,12,20,.....,100$? Find its sum.

Answer 1

Hint: In order to find the type of the sequence, whether the series is in Arithmetic Progression (AP) or Geometric position (GP), or any other series. If the series is found out to be an AP, then find the common difference, and substitute all the values given in the formula of $n$ th term of an AP, and sum of $n$ terms in an AP , then evaluate and find the number of terms and sum of terms.

Formula used:
${T_n} = a + \left( {n - 1} \right)d$
${S_n} = \dfrac{n}{2}\left( {a + l} \right)$
${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$

Complete step by step solution:
We are given the series $4,12,20,.....,100$. Where the first term is ${T_1} = 4$, second term is ${T_2} = 12$, third term is ${T_3} = 20$, and similarly the last term becomes the $n$ th term or ${T_n} = 100$ .
To find the type of series, we need to find the common difference, if it's the same for all, then it’s an AP (Arithmetic Progression).
We know that $d$, the common difference is the difference between two consecutive terms in an AP, which can be found out by subtracting any ${T_n}$ term from ${T_{n + 1}}$ term.
So, subtracting ${T_1}$ from ${T_2}$, and we get:
${T_2} - {T_1} = 12 - 4 = 8$
So, $d = 8$
Similarly, subtracting ${T_2}$ from ${T_3}$, and we get:
${T_3} - {T_2} = 20 - 12 = 8$
So, $d = 8$
Since, both the differences are equal, that means the series is in AP, and it’s our common difference $d = 8$.

From the Arithmetic Progression series, we know that to find $n$th term, we use the formula ${T_n} = a + \left( {n - 1} \right)d$, where \[a\] is nothing but the first term that is $a = 4$, and $n$ is the number of terms and $d$ is the common difference.
Since, we are given the values of ${T_n}$, $d$ and $a$. We need to find the value of $n$.
Substituting the values of $a$,${T_n}$ and $d$ in the equation ${T_n} = a + \left( {n - 1} \right)d$:
$
  {T_n} = a + \left( {n - 1} \right)d \\
   \Rightarrow 100 = 4 + \left( {n - 1} \right)8 \\
 $


Opening the parenthesis in the above equation:
$
  100 = 4 + \left( {n - 1} \right)8 \\
   \Rightarrow 100 = 4 + 8n - 8 \\
   \Rightarrow 100 = 8n - 4 \\
 $
Adding both the sides by $4$, and we get:
$
  100 = 8n - 4 \\
   \Rightarrow 100 + 4 = 8n - 4 + 4 \\
   \Rightarrow 104 = 8n \\
 $
Dividing both the sides by $8$, in order to get the value of $n$:
$
  8n = 104 \\
   \Rightarrow \dfrac{{8n}}{8} = \dfrac{{104}}{8} \\
   \Rightarrow n = 13 \\
 $

Therefore, the number of terms, in the AP $4,12,20,.....,100$ is $13$.

Since, the series is in AP, so we know that for an AP, the sum of the terms is calculated by ${S_n} = \dfrac{n}{2}\left( {a + l} \right)$, where $a$ is the first term that is $a = 4$, $l$ is the last term that is $l = 100$ , $n$ is the number of terms that we found $n = 13$, and ${S_n}$ is the sum of $n$ terms, which we need to find out.

Substituting $a$,$l$,$n$ in the equation ${S_n} = \dfrac{n}{2}\left( {a + l} \right)$:
$
  {S_n} = \dfrac{n}{2}\left( {a + l} \right) \\
   \Rightarrow {S_n} = \dfrac{{13}}{2}\left( {4 + 100} \right) \\
 $
Opening the parenthesis and solving it further and we get:
$
  {S_n} = \dfrac{{13}}{2}\left( {4 + 100} \right) \\
   \Rightarrow {S_n} = \dfrac{{13}}{2}\left( {104} \right) \\
   \Rightarrow {S_n} = \dfrac{{13 \times 104}}{2} = 13 \times 52 \\
   \Rightarrow {S_n} = 676 \\
 $
Therefore, the sum of all the terms of the sequence $4,12,20,.....,100$ is $676$.

Note:
Since, we got to know that the series was in AP, if it was not then we would have checked for GP, and for that we would have divided the second term by first and third term by second, if those two give the same results, then it’s in GP. If not GP, then we will check for another series.
We can use this ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ formula if the last term is not known.