Question

Number of terms in the sequence $1,3,6,10,15,...,5050$is
A. $50$
B. $75$
C. $100$
D. $125$

Answer 1

Hint: First, we shall analyze the given information so that we can able to solve the problem. Here in this question, we are given a sequence$1,3,6,10,15,...,5050$and we need to calculate the number of terms in the given sequence. First, we need to find the sum of the given sequence and then we need to apply the formula to find the required answer.
Formula to be used:
a) ${T_n} - {T_{n - 1}} = n$ where $n$is the number of terms, $T$ is the ${n^{th}}$ term of the sequence.
b) The formula of the sum of the $n$ terms is as follows.
$1 + 2 + 3 + 4 + 5 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}$

Complete step by step answer:
    We are given a sequence$1,3,6,10,15,...,5050$.
Here, we need to calculate the number of terms in the given sequence.
Let us consider the number of terms as$n$ .
Let us assume that${T_n} = 5050$ and ${S_n}$ be the sum of the given sequence.
Hence, we get${S_n} = 1 + 3 + 6 + 10 + 15 + ... + {T_n}$ …………$\left( 1 \right)$
Now, we shall calculate the sum of the sequence${S_{n - 1}}$
Thus, we have${S_{n - 1}} = 1 + 3 + 6 + 10 + 15 + ... + {T_{n - 1}}$ …………$\left( 2 \right)$
Now, from$\left( 1 \right)$we get${S_n} - {T_n} = 1 + 3 + 6 + 10 + 15 + ... + {T_{n - 1}}$
Now we shall apply $\left( 2 \right)$in the above equation.
Thus, ${S_n} - {T_n} = {S_{n - 1}}$
$ \Rightarrow {T_n} = {S_n} - {S_{n - 1}}$
So, we need to subtract$\left( 1 \right)$from$\left( 2 \right)$.
Here, we shall shift one term while subtracting the terms.
${T_n} = {S_n} - {S_{n - 1}}$
       $ = 1 + 3 - 1 + 6 - 3 + 10 - 6 + 15 - 10 + ... + {T_n}{\text{ }} - {\text{ }}{T_{n - 1}}$
       $ = 1 + 2 + 3 + 4 + 5 + ... + n$ (${T_n} - {T_{n - 1}} = n$)
We need to apply the formula of the sum of the $n$ terms in the above equation.
$1 + 2 + 3 + 4 + 5 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}$
Hence, we get${T_n} = \dfrac{{n\left( {n + 1} \right)}}{2}$
Since${T_n} = 5050$we shall compare both equations.
$ \Rightarrow 5050 = \dfrac{{n\left( {n + 1} \right)}}{2}$
$ \Rightarrow 5050 \times 2 = n\left( {n + 1} \right)$
$ \Rightarrow {n^2} + n = 10100$
$ \Rightarrow {n^2} + n - 10100 = 0$
Now, we shall split the middle term as follows.
$ \Rightarrow {n^2} + 101n - 100n - 10100 = 0$
We need to pick the common terms.
$ \Rightarrow n\left( {n + 101} \right) - 100\left( {n + 101} \right) = 0$
$ \Rightarrow \left( {n - 101} \right)\left( {n + 101} \right) = 0$
Hence, we get$n = 100$ or$n = - 101$
Since the number of terms cannot be negative, $n = - 101$is not possible.
Hence, $n = 100$.

So, the correct answer is “Option C”.

Note: Let us consider the number of terms as$n$ . To obtain the required number of terms, we have compared${T_n}$ . Here, we found$n = 100$ or$n = - 101$.
          And, the number of terms must be positive. So, we neglect$n = - 101$.
         Therefore, we get the required number of terms$n = 100$.