Latest Elitmus Aptitude Question SOLUTION: what is the sum of these 1230 sequences such as 1,2,1,2,2,1,2,2,2,1,2,2,2,2.....

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(#M40037745) ELITMUS Numerical Ability question Keep an EYE

what is the sum of these 1230 sequences such as 1,2,1,2,2,1,2,2,2,1,2,2,2,2.....

Asked In Elitmus naina (9 years ago)

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Saraswathy  (9 years ago)

Dont know wats wrong... couldnt able to post full solution...
1,2,1,2,2,1,2,2,2,1,2,2,2,2....
We have to find the no. of 1's nd no. of 2's in 1230 terms...
Lets take 1st 2 terms ... 1, 2
no. of 1's is 1 nd no. of 2's is 1 => (1+1)
1st 5 terms ... 1,2,1,2,2
=> (1+1)+(1+2)
1st 9 terms ... 1,2,1,2,2,1,2,2,2
=> (1+1)+(1+2)+(1+3)
=> (1+1+1)+(1+2+3)
3 + 3(3+1)/2 ... [Its of the form n + n(n+1)/2]...
So in 1st 9 terms, no. of 1's is "n", no. of 2's is n(n+1)/2...
So sum of 1st 9 terms ll be n*1 + n(n+1)*2/2
=> 3*1 + 3(4)*2/2 => 15...
Provided n + n(n+1)/2
So for the 1230 terms,
n + n(n+1)/2
n=48
48 + 48*49/2 = 1224 which is 1+2+2+2+2+2 = 11
So total sum ll be = 48*1 + 1176*2 + 11 = 2411
Ans : 2411

Is this solution Helpfull? Yes (23) | No (1)  

Hrishabh Sobti  (9 years ago)

The series goes like :
1 2
1 2 2
1 2 2 2
1 2 2 2 2
1 2 2 2 2 2
1 2 2 2 2 2 2...
and so on

we see that for nth 1 there are n number of 2s
=> given 'n' 1s, there are (n/2)[2a+(n-1)d]_____formula for sum of AP
lets see
for 50 1s, there are 1275 2s => total number of terms = 50+1275 which is >1230
for 49 1s, there are 1200 2s => total number of terms = 49+1200=1249 which is >1230, but we can ignore the surplus 2s (since the last 49 terms are 2)
extra terms = 19 (=1249-1230)
Therefore numberof 2s = 1200-19=1181
=> (49*1) + (1181*2) = 2411

Is this solution Helpfull? Yes (18) | No   

Saraswathy  (9 years ago)

This is my complete solution...

1,2,1,2,2,1,2,2,2,1,2,2,2,2....

We have to find the no. of 1's nd no. of 2's in 1230 terms...

Lets take 1st 2 terms ... 1, 2
no. of 1's is 1 nd no. of 2's is 1 => (1+1)
1st 5 terms ... 1,2,1,2,2
=> (1+1)+(1+2)
1st 9 terms ... 1,2,1,2,2,1,2,2,2
=> (1+1)+(1+2)+(1+3)
=> (1+1+1)+(1+2+3)
3 + 3(3+1)/2 ... [Its of the form n + n(n+1)/2]...
So in 1st 9 terms, no. of 1's is "n", no. of 2's is n(n+1)/2...
So sum of 1st 9 terms ll be n*1 + n(n+1)*2/2
=> 3*1 + 3(4)*2/2 => 15...
Provided n + n(n+1)/2

So for the 1230 terms,
n + n(n+1)/2

Is this solution Helpfull? Yes (7) | No (1)  

Saraswathy  (9 years ago)

So for the 1230 terms,
n + n(n+1)/2
n=48
48 + 48*49/2 = 1224 which is close to 1230
no. of 1's in 1224 terms = 48
no. of 1's in 1224 terms = 48*49/2 = 1176
So the next 6 terms after 1224 ll be 1,2,2,2,2,2 => sum = 1+2+2+2+2+2 = 11
So total sum ll be = 48*1 + 1176*2 + 11 = 2411
Ans : 2411

Is this solution Helpfull? Yes (5) | No   

Saraswathy  (9 years ago)

1,2,1,2,2,1,2,2,2,1,2,2,2,2....

We have to find the no. of 1's nd no. of 2's in 1230 terms...

Lets take 1st 2 terms ... 1, 2
no. of 1's is 1 nd no. of 2's is 1 => (1+1)
1st 5 terms ... 1,2,1,2,2
=> (1+1)+(1+2)
1st 9 terms ... 1,2,1,2,2,1,2,2,2
=> (1+1)+(1+2)+(1+3)
=> (1+1+1)+(1+2+3)
3 + 3(3+1)/2 ... [Its of the form n + n(n+1)/2]...
So in 1st 9 terms, no. of 1's is "n", no. of 2's is n(n+1)/2...
So sum of 1st 9 terms ll be n*1 + n(n+1)*2/2
=> 3*1 + 3(4)*2/2 => 15...
Provided n + n(n+1)/2

Is this solution Helpfull? Yes (2) | No (1)  

LOKESH SAMANT  (9 years ago)

1230 terms contain 49 1's and remaining 2's
so 1181*2+49 = 2411

Is this solution Helpfull? Yes (2) | No (1)  

M.Prasanth  (9 years ago)

can u plss explain hw n=48 came saraswathy???

Is this solution Helpfull? Yes (2) | No   

naina  (9 years ago)

@ SARASWATHY
i m not able to understand hw u r taking no of terms plz explain 1st 5 terms ... 1,2,1,2,2
=> (1+1)+(1+2) why r u taking 1+2???
@ LOKI
can u plz explain hw r u calculating no.of 1' s & 2's???

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Saraswathy  (9 years ago)

I could able to post my complete solution...
Ans ll be 2411
1,2,1,2,2,1,2,2,2
(1+1)+(1+2)+(1+3)
here (1+1) represents no. of 1's is 1 nd no. of 2's is 1...
(1+2) represents no. of 1's is 1 nd no. of 2's is 2...
(1+3) represents no. of 1's is 1 nd no. of 2's is 3...
(1+1)+(1+2)+(1+3)
We can write as (1+1+1)+(1+2+3)
3 + sum of 3 terms => 3+ 3(4)/2 => 9
So in first 9 terms no. of 1's is 3 nd no. of 2's is 6...
So sum ll be 3*1 + 6*2 = 15

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tirupathi reddy  (9 years ago)

I guess 2412
No.of 1's are 48 and remain 2's I.e., 1182
1182*2+48*1=2412.
Can anyone verify it....

Is this solution Helpfull? Yes | No (1)  

Abhishek Sa  (9 years ago)

they said sequence not terms

Is this solution Helpfull? Yes | No (2)  

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