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Question: Number of terms in the sequence \(1,3,6,10,15,...,5050\)is A. \(50\) B. \(75\) C. \(100\) ...

Number of terms in the sequence 1,3,6,10,15,...,50501,3,6,10,15,...,5050is
A. 5050
B. 7575
C. 100100
D. 125125

Explanation

Solution

First, we shall analyze the given information so that we can able to solve the problem. Here in this question, we are given a sequence1,3,6,10,15,...,50501,3,6,10,15,...,5050and 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) TnTn1=n{T_n} - {T_{n - 1}} = n where nnis the number of terms, TT is the nth{n^{th}} term of the sequence.
b) The formula of the sum of the nn terms is as follows.
1+2+3+4+5+...+n=n(n+1)21 + 2 + 3 + 4 + 5 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}

Complete step by step answer:
We are given a sequence1,3,6,10,15,...,50501,3,6,10,15,...,5050.
Here, we need to calculate the number of terms in the given sequence.
Let us consider the number of terms asnn .
Let us assume thatTn=5050{T_n} = 5050 and Sn{S_n} be the sum of the given sequence.
Hence, we getSn=1+3+6+10+15+...+Tn{S_n} = 1 + 3 + 6 + 10 + 15 + ... + {T_n} …………(1)\left( 1 \right)
Now, we shall calculate the sum of the sequenceSn1{S_{n - 1}}
Thus, we haveSn1=1+3+6+10+15+...+Tn1{S_{n - 1}} = 1 + 3 + 6 + 10 + 15 + ... + {T_{n - 1}} …………(2)\left( 2 \right)
Now, from(1)\left( 1 \right)we getSnTn=1+3+6+10+15+...+Tn1{S_n} - {T_n} = 1 + 3 + 6 + 10 + 15 + ... + {T_{n - 1}}
Now we shall apply (2)\left( 2 \right)in the above equation.
Thus, SnTn=Sn1{S_n} - {T_n} = {S_{n - 1}}
Tn=SnSn1\Rightarrow {T_n} = {S_n} - {S_{n - 1}}
So, we need to subtract(1)\left( 1 \right)from(2)\left( 2 \right).
Here, we shall shift one term while subtracting the terms.
Tn=SnSn1{T_n} = {S_n} - {S_{n - 1}}
=1+31+63+106+1510+...+Tn  Tn1= 1 + 3 - 1 + 6 - 3 + 10 - 6 + 15 - 10 + ... + {T_n}{\text{ }} - {\text{ }}{T_{n - 1}}
=1+2+3+4+5+...+n= 1 + 2 + 3 + 4 + 5 + ... + n (TnTn1=n{T_n} - {T_{n - 1}} = n)
We need to apply the formula of the sum of the nn terms in the above equation.
1+2+3+4+5+...+n=n(n+1)21 + 2 + 3 + 4 + 5 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}
Hence, we getTn=n(n+1)2{T_n} = \dfrac{{n\left( {n + 1} \right)}}{2}
SinceTn=5050{T_n} = 5050we shall compare both equations.
5050=n(n+1)2\Rightarrow 5050 = \dfrac{{n\left( {n + 1} \right)}}{2}
5050×2=n(n+1)\Rightarrow 5050 \times 2 = n\left( {n + 1} \right)
n2+n=10100\Rightarrow {n^2} + n = 10100
n2+n10100=0\Rightarrow {n^2} + n - 10100 = 0
Now, we shall split the middle term as follows.
n2+101n100n10100=0\Rightarrow {n^2} + 101n - 100n - 10100 = 0
We need to pick the common terms.
n(n+101)100(n+101)=0\Rightarrow n\left( {n + 101} \right) - 100\left( {n + 101} \right) = 0
(n101)(n+101)=0\Rightarrow \left( {n - 101} \right)\left( {n + 101} \right) = 0
Hence, we getn=100n = 100 orn=101n = - 101
Since the number of terms cannot be negative, n=101n = - 101is not possible.
Hence, n=100n = 100.

So, the correct answer is “Option C”.

Note: Let us consider the number of terms asnn . To obtain the required number of terms, we have comparedTn{T_n} . Here, we foundn=100n = 100 orn=101n = - 101.
And, the number of terms must be positive. So, we neglectn=101n = - 101.
Therefore, we get the required number of termsn=100n = 100.