Latest Elitmus Aptitude Question SOLUTION: 1,2,1,2,2,1,2,2,2,1,2,2,2,2,1......................... Find the Sum of 1230 terms of the series..?

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(#M40120740) ELITMUS Numerical Ability question Quant Keep an EYE

1,2,1,2,2,1,2,2,2,1,2,2,2,2,1.........................
Find the Sum of 1230 terms of the series..?

Asked In Elitmus MOHAMMAD AMIR (9 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

Rajmohan  (9 years ago)

series =1,2,1,2,2,1,2,2,2
first term 1
second term 2,1
third term 2,2,1
...so on
no of terms is 1+2+3+4+.....
after checking the sum almost equal to 1230 is 49 terms (you can check 49*(49+1)/2=1225)
and from given series 1+(2+1)+(2+2+1)+(2+2+2+1)+.....(1+2*49)
sum is 1+3+5+7.......97 it means sum of 97 odd nos.
=49*49=2401
last 5 terms (2+2+2+2+2) part of 50th terms 2,2,2,2,2...........1
ans is 2401+10=2411

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aman aditya  (9 years ago)

ans>2411

each series has one more term then previous
i.e
1,2 ..2term
1,2,2...3 term
so for 1230 term there will be 48 complete series
and last six term will be 1,2,2,2,2,2,

now each of the series addition is 2 more then previous series
i.e
1+2=3
1+2+2=5
so on
so its form an A.P
so sum of 48 term will be
S=n/2(2*a+(n-1)d)
48/2(2*3+47*2)
=2400
now
sum of remaining last eight term
1,2,2,2,2,2=11
hence answer is 2400+15=2411
Read more at http://www.m4maths.com/placement-puzzles.php#qloSY1L2GlStYzTo.99

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shiv  (9 years ago)

ans=2421
1,2,1,2,2,1,2,2,2,1,2,2,2,2,...................
we are going to pair the number firstly as
first term =1
second term =2,1
third term =2,2,1
fourth term=2,2,2,1
in first term no of term =1
in second term no of term =2.............
so
1+2+3+4.............
we have to check whether this sum is approximately equal to 1230
so by taking 49 terms we are getting sum as 1225
in all the terms we have a unity.
and from second terms we have (n-1) numbers of .
so sum=49+(2+4+6+8+.............98)=2411
but we tis is the sum of 1225 terms only
next five terms =2,2,2,2,2
sum of these terms=10
total sum=2421

Is this solution Helpfull? Yes (4) | No (4)  

siddharth prakash singh  (9 years ago)

series is 1,2,1,2,2,1,2,2,2
first term 1
second term 2,1
third term 2,2,1
fourth term 2,2,2,1
and so on......
thus no of terms= 1+2+3+4+........
we have to calculate the sum of 1230 terms.
after hit nd trial we will find sum of 49 terms is 1225. after that there will be 2,2,2,2,2.(as we require 1230 terms hence (1230-1225)=5 term of next sequence.)
now,
sum of upto 1st term is 1 (i.e- 1)
sum of upto 2nd term is 4 (i.e- 1+(2,1))
sum of upto 3rd term is 9 (i.e- 1+(2,1)+(2+2+1))
sum of upto 4th term is 16 (i.e- 1+(2,1)+(2+2+1)+(2+2+2+1))
sum of upto 5th term is 25 (i.e- 1+(2,1)+(2+2+1)+(2+2+2+1)+(2+2+2+2+1)) and so on....
hence sum of upto 49 terms will be 49*49 =2401
so required sum =2401*(2+2+2+2+2)=2411

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Siddharth jain  (9 years ago)

there would be in total 48 complete series and 6 extra terms which include 1,2,2,2,2,2

so suum of 48 tems is equal to 2400
and plus 6 terms that is 11
so ans is 2411

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aman aditya  (9 years ago)

ans>2415

each series has one more term then previous
i.e
1,2 ..2term
1,2,2...3 term
so for 1230 term there will be 48 complete series
and last eight term will be 1,2,2,2,2,2,2,2

now each of the series addition is 2 more then previous series
i.e
1+2=3
1+2+2=5
so on
so its form an A.P
so sum of 48 term will be
S=n/2(2*a+(n-1)d)
48/2(2*3+47*2)
=2400
now
sum of remaining last eight term
1,2,2,2,2,2,2,2=15
hence answer is 2400+15=2415

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