Let Tk denote the kth term of an arithmetic progression. Suppose th

Question

GMATBusters’ Quant Quiz Question -6

Let Tk denote the kth term of an arithmetic progression. Suppose there are positive integers m≠n such that Tm =  \frac{1}{n}  and Tn =  \frac{1}{m} . Then Tm*n equals 
(a)  \frac{1}{mn } 
(b)  \frac{1}{m}+\frac{1}{n } 
(c) -1
(d) 0
(e) 1

ashwini2k6jha · Answer

given Tm=1/n and Tn=1/m,
Let a be first number of AP series and common diff=d
we have to find Tmn=a+(mn-1)d=??
Tm=a+(m-1)d=1/n ------1
Tn=a +(n-1)d=1/m-------2
substract 2from 1
d=1/mn
from 1 get a by putting value of d, we get a=1/mn
Now Tmn= 1/mn+(mn-1)x1/mn=1 
Answer D

Raxit85 · Answer

Let, common difference=d

rth term =tr
1st term= a
Now mth term= tm=a+(m−1)d
Since, tm=n1
n1=a+(m−1) …… (1)
Now nth term tn=a+(m−1)d
Given , tn=m1
m1=a+(n−1) …… (2)
Let's subtract (2) from (1)
n1−m1=(m−1)d−(n−1)d
mnm−n=(m−n)d
d=mn1
Let's put value d in equation (1)
n1=a+(m−1)mn1
n1=a+mnm+mn1
n1=a+n1−mn1
n1=a+n1−mn1
a=mn1
Now mnth term ,
tmn=mn1+(mn−1)mn1
=mn1+1−mn1
tmn=1
Hence, Ans. E

gmatbuster , the option numbering need to correct.

Kinshook · Answer

Given: 
1. Let Tk denote the kth term of an arithmetic progression. 
2. Suppose there are positive integers m≠n such that Tm = 1/n and Tn = 1/m

Asked: Then Tm*n equals 
(a) 1/mn
(b) 1/m+1/n
(c) -1
(d) 0
(e) 1

If a is the first term of the arithmetic progression and d be the common difference.

Tk = a + (k-1)d

Tm = a + (m-1)d = 1/n
Tn = a + (n-1)d = 1/m

(m-n)d = 1/n - 1/m = (m-n)/mn
d = 1/mn

a = 1/n - (m-1)/mn = {m - (m-1)} / mn = 1/mn

Tm*n = a + (mn-1)d = 1/mn + (mn-1)/mn = 1/mn + 1 - 1/mn = 1

IMO E

techwave · Answer

Looks like Typo for answers, C printed twice.

Answer : E (1)

Let' 'd' is the common difference of successive members.

a is the initial term.

There fore per prompt.

a+(m–1)d= 1/n ---(1)

a+(n–1)d= 1/m----(2)

Substracting 1- 2

(m–n)d= 1/n- 1/m
d=1/(mn)

Substituting d in (1) gives a =1/mn
now we have both a and d and a=d

To calculate:
Tm*n=a+(mn–1)d
=1/mn + (mn-1)/mn
=1/mn + 1 -1/mn = 1

Archit3110 · Answer

for given AP we can write
Tm= a+(m-1)*d
Tn=a+(n-1)*d
1/n= a+(m-1)*d---(1)
1/m=a+(n-1)*d--(2)
subtract 1 from 2
m-n/mn = m-n
d=1/mn
substitute in (1)
we get 
a+(m-1)*1/mn=1/n
a = 1/mn
so 
Tmn = a +(mn-1)*d
Tmn= 1/mn+(mn-1)*1/mn
Tmn = 1/mn +1-1mn
Tmn = 1
IMO E (1)

Let Tk denote the kth term of an arithmetic progression. Suppose there are positive integers m≠n such that Tm = 1/n and Tn = 1/m. Then Tm*n equals
(a) 1/mn
(b) 1/m+1/n
(c) -1
(d) 0
(e) 1

shameekv1989 · Answer

Let Tk denote the kth term of an arithmetic progression. Suppose there are positive integers m≠n such that Tm = 1/n and Tn = 1/m. Then Tm*n equals
(a) 1/mn
(b) 1/m+1/n
(c) -1
(c) 0
(d) 1

T_k = a+2(k-1)d

T_m = a+2(m-1)d 
 T_n = a+2(n-1)d

T_m - T_n = 2md - 2d - 2nd + 2d = 2md - 2nd

\frac{1}{n} - \frac{1}{m} = 2d(m-n)

\frac{m - n}{mn} = 2d(m-n)

d = \frac{1}{2mn}

T_m = a+2(m-1)d  =>  \frac{1}{n} = a+\frac{2(m-1)}{2mn}

a= \frac{1}{n}-\frac{(m-1)}{mn}

a= \frac{(m-m+1)}{mn} = \frac{1}{mn}

T_{mn} = a+2(mn-1)d

T_{mn} = \frac{1}{mn}+\frac{2(mn-1)}{2mn}

T_{mn} = \frac{1+mn-1}{mn}  =  \frac{mn}{mn}

Answer - D

bM22 · Answer

Let us assume Tm = T3 and Tn = T4 for quick solution

=> T3 = 1/4 & T4 = 1/3
=> common diff d = 1/12

so now a+2d = 1/4 so a = 1/4 - 1/6 = 1/12

so now T12 = a+ 11d => 1/12 + 11(1/12) = 1.

So in general, for any Tmn, the value will be 1.

Answer D

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Let Tk denote the kth term of an arithmetic progression. Suppose th

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