For each positive integer n, the nth positive triangular number is equ

Question

GMATH  practice exercise (Quant Class 17)

https://i.ibb.co/qjVrNkW/24-Feb19-4z.gif

For each positive integer n, the nth positive triangular number is equal to the number of dots in the triangular arrangement with n dots on a side. The first positive triangular numbers are 1, 3, 6, 10, 15, and 21, as shown. Which of the following numbers is NOT a positive triangular number?

(A) 105
(B) 210
(C) 300
(D) 311
(E) 378

CharlieharpeR · Answer

So study these numbers carefully. You will fibd that T(n)= Sum of all natural numbers from 1 to n. We know that this sum is given by the general formula (n(n+1))/2.

So one method can be assuming that each of these options is given by such a formula and trying to find a value of n that satisfies it. The one for which this is not possible is your answer. Right now i can’t think of any other method off the cuff. So let me fasten up the above method for you. I’ll update here if i think of some other method.

Assume each option = n(n+1)/2
So 2* each option = product of 2 consecutive nunbers. Let’s consider each option now.

As you browse the options you’ll find that 311 is prime so 311*2 can never be expressed as the sume of 2 consecutive numbers so the answer is D

Cheers

Posted from my mobile device

Archit3110 · Answer

values of the triangular no are given n*(n+1)/2 
so out of given options 311 is prime no. which is not possible
IMO D

fskilnik · Answer

{T_n} = 1 + 2 + \ldots + n\,\,\,\,\left( {n \ge 1\,\,{\mathop{\rm int}} } \right)

{T_n} = {{n \cdot \left( {n + 1} \right)} \over 2}\,\,\,\,\,\,\left

?\,\,\,:\,\,\,\underline {{\rm{not}}} \,\,\,T

\left( A \right)\,\,\left\{ \matrix{
 \,n \cdot \left( {n + 1} \right) = 2 \cdot 105 = 210 \hfill \cr 
 \,15 \cdot 15 = 225 \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{try}}!} \,\,\,\,{{14 \cdot 15} \over 2} = 7 \cdot 15 = 105\,\,\,{\rm{works}}!\,\,\,\,\, \Rightarrow \,\,\,\,\,105 = {T_{14}}

\left( B \right)\,\,\left\{ \matrix{
 \,n \cdot \left( {n + 1} \right) = 2 \cdot 210 = 420 \hfill \cr 
 \,20 \cdot 20 = 400 \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{try}}!} \,\,\,\,{{20 \cdot 21} \over 2} = 10 \cdot 21 = 210\,\,\,{\rm{works}}!\,\,\,\,\, \Rightarrow \,\,\,\,\,210 = {T_{20}}

\left( C \right)\,\,\left\{ \matrix{
 n \cdot \left( {n + 1} \right) = 2 \cdot 300 = 600 \hfill \cr 
 25 \cdot 25 = 625 \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{try}}!} \,\,\,\,{{24 \cdot 25} \over 2} = 12 \cdot 25 = 300\,\,\,{\rm{works}}!\,\,\,\,\, \Rightarrow \,\,\,\,\,300 = {T_{24}}

\left( D \right)\,\,\,{T_{24}} < 311 < 300 + 25 = {T_{24}} + 25 = {T_{25}}

The correct answer is (D).

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

Kinshook · Answer

fskilnik

GMATH  practice exercise (Quant Class 17)

https://i.ibb.co/qjVrNkW/24-Feb19-4z.gif

For each positive integer n, the nth positive triangular number is equal to the number of dots in the triangular arrangement with n dots on a side. The first positive triangular numbers are 1, 3, 6, 10, 15, and 21, as shown. Which of the following numbers is NOT a positive triangular number?

Tn = 1 + 2 + 3 + 4 + ...... + Tn-1 = n(n+1)/2

Every term can be written in the form of n(n+1)/2

(A) 105: n(n+1)/2 = 105; n^2+n-210 = 0; (n+15)(n-14) = 0; n = 14; Possible
(B) 210: n(n+1)/2 = 210; n^2+n-420 = 0; (n+21)(n-20) = 0; n = 20; Possible
(C) 300: n(n+1)/2 = 300; n^2+n-600 = 0; (n+25)(n-24) = 0; n = 24; Possible
(D) 311: n(n+1)/2 = 311; n^2+n-622 = 0; Can not be written in the form n(n+1)/2; Not Possible
(E) 378: n(n+1)/2 = 378; n^2+n-756 = 0; (n+28)(n-27) = 0; n = 27; Possible

IMO D

bumpbot · Answer

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For each positive integer n, the nth positive triangular number is equ

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