TCS Series and Progressions-Quiz-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 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 = 1/n
Let, Ti denote the ith term, Si denote the sum upto ith term.
S1=T1=1/1=1.

S2=1/2
hence, T2=S2-S1=1/2 - 1 = -1/2

S3=1/3
hence, T3=S3-S2= 1/3 - 1/2 = -1/6

similarly, T4= -1/12 and so on.

hence, for 4 terms, product = (1) (-1/2) (-1/6) (-1/12) = (-1/144) = -1/(3! * 4!)

hence by observing, option -1/(2013! * 2014!) is correct.

1st term gives 2nd and 3rd term. .
1st term* 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*2=30th term
so that 156*5=780(30th term)
31st term is 780+1=781

IF WE TAKE hcf 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*100 =110
110 = 89 + 0.15n
solving this
n = 140 days
which is 20th of may

(1,3,5,..1999) here n=1000
s1=n/2(2a+(n-1)d)=10000000
(-2,-4,-6...,-2000) here n=1000
s2=-1001000
so avg term= s1+s2/2000term
=-1000/2000
=-1/2 ans.

Text solution:
11+12*1=23
23+12*2=47
47+12*3=83
83+12*4=131
131+12*5=191

Video solution:

b)1
1,2,3,4(1-1time 2- 2times 3-3 4-4)=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 n(n+1)/2
10.(21.22)/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