TCS NQT Series and Progressions Questions and Answers Quiz 1

1,2,3,4(1-1time 2- 2times 3-3times 4-4times)=10 terms completes cycle and starts from 1
1,2,3,4(1-2 2-4 3-6 4-8)= 20 terms completes cycle and starts
from 1
(1,2,3,4)each digit 3 time to its value =30 terms completes cycle and starts from 1

10+20+30+40+50+.......=x
x is nearer value to 2320 solving \frac{n(n+1)}{2}
10 \times \frac{(21\times 22)}{2}= 2310
analysing it 2310 completes cycle and starts from 1 again
now it 22 times
(1-22 times 2-44 times ......)
2320 position will occupied by 1

AERDOI

Having AD as first 2 letters we can have 24 words.

Then AED we have 6 words, ie 24+6=30.

Having AEI again 6 words, ie 30+6= 36,
Having AEO, 36+6=42.

AERD we have 2 words, AERDIO and AERDOI.

Since in alphabetical order, answer is AERDOI

1234 => 4 times
12344 => 5 times
123444 => 6 times
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4 + 5 + 6 + ...n < 200

\frac{n(n+1)}{2} - 6 < 200

So n = 19 i.e. 4+5+6+...19 = 184

After the 184th term,the sequence has 20 terms 12344444444444444444 = 20 terms => 200th term = 4

t2 + t19 = t8 + t15 + t12

=> a+d + a+18d = a+7d + a+14d + a+11d

=> a + 13d = 0

=> t14 = 0

i.e. 14th term is zero.

Sn = \frac{1}{n}
Let, Ti denote the ith term, Si denote the sum upto ith term.
S1=T1= \frac{1}{1} = 1

S2= \frac{1}{2}
hence, T2=S2-S1= \frac{1}{2}  - 1 = \frac{-1}{2}

S3= \frac{1}{3}
hence, T3=S3-S2= \frac{1}{3} - \frac{1}{2} = \frac{-1}{6}

similarly, T4= \frac{-1}{12} and so on.

Hence, for 4 terms, product = (1) \frac{-1}{2} \frac{-1}{6} \frac{-1}{12}

= \frac{-1}{144}

= \frac{-1}{3 ! \times 4 ! }

hence by observing, option \frac{-1} {2013 !  \times 2014 ! } is correct.

1st term gives 2nd and 3rd term. .
1st term x 5= 5(2nd term)
2nd term +1=6(3rd term)
multiply current position by 2 gives the position of term which is multiplied by no 5
2nd term gives 4th and 5th term. .
likewise check for others
15th term gives 30th and 31st term. .
ans:15th term x 2=30th term
so that 156 x 5=780(30th term)
31st term is 780+1=781

IF WE TAKE LCM of 9 and 33 then it cames 99
so after 99 sec they will flash together.
in 99 sec red flash 11 times and yellow flash 3 times.
next 99 sec same will be repeat.
and in remaining 52 sec red flash 5 times and yellow flash 1 time
so Total red flash is 27 and 7 times
total difference is 20

20 may
100 + 0.1 x 100 =110
110 = 89 + 0.15n
solving this
n = 140 days
which is 20th of may

(1,3,5,..1999) here n=1000

sum =\frac{n}{2}(a_{1}+ a_{n})=1000000
(-2,-4,-6...,-2000) here n=1000
sum =\frac{n}{2}(a_{1}+ a_{n})= - 1001000
so avg term= \frac{1000000+(-1001000)}{2000} term
=\frac{-1000}{2000}
= \frac{-1}{2} ans.