Latest Elitmus Aptitude Question SOLUTION: 1 2 1 2 2 1 2 2 2 1 2 2 2 2 1 2 ...... sum of the series upto 1230 terms

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(#M40032561) ELITMUS Numerical Ability question Apti Keep an EYE

1 2 1 2 2 1 2 2 2 1 2 2 2 2 1 2 ...... sum of the series upto 1230 terms

Asked In Elitmus Arnab Halder (10 years ago)

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RAKESH  (10 years ago)

1 2 1 2 2 1 2 2 2 1 2 2 2 2 1 2 ......
=(1+2),(1+2+2),(1+2+2+2),(1+2+2+2+2)......
if N be no. of terms then
N=(1+1)+(1+2)+(1+3)+(1+4)+...
N= n+ n(n+1)/2 = (n^2+3n)/2
now (n^2+3n)/2 = 1230 => n^2+3n=2460
if n=48, (n^2+3n)=2448 & N=2448/2=1224
so series will be
sum of (1224 term + 6 term)
[(1+2)+(1+2+2)+(1+2+2+2)+(1+2+2+2+2)+...+(1+2+2...48 times 2)]+(1+2+2+2+2+2)
so no. of 1's in series=49
no. of 2's in series=(1230-49)=1181

sum of series= 49*1+1181*2 = 49+2362=2411

ans 2411

Is this solution Helpfull? Yes (51) | No (2)  

Neeraj kumar jha  (10 years ago)

1 2
1 2 2
1 2 2 2
1 2 2 2 2
so on, it means at first for 1st 1 we have one 2, for second 1 total num of twos are 1+2, for 3rd 1 total num of twos are 1+2+3.....so on. so if we take 48 one then total num of twos are 1+2+...+48 = 48*49/2 = 1176 twos. So total num of terms are 48 ones and 1176 tows = 1124 terms ...again after it one 1 comes it becomes 1125th term and after that 5 tows ...so total becomes 1130 terms. So sum is 48*1+1176*2+1+10 = 2411

Is this solution Helpfull? Yes (12) | No (1)  

PRATIM JANA  (8 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 (5) | No   

Avigyan  (10 years ago)

No of terms: Sums
1 1 1
2 21 3
3 221 5
4 2221 7
5 22221 9
=15 terms total total sums=25
and so on..
if we calculate the no.of terms column upto 49no. that will gives us the total no.of terms upto 49no. that will be equal to n(n+1)/2 = 49*25=1225.
so, the total sum for 1225 terms will be equal to 49/2(2*1+(49-1)*2)=2401 (since the series in sum column is 1 3 5 7 9 ....)
now the next 5 terms will be like this... 2 2 2 2 2 (can see the pattern)
so the total sums will be (2*5+2401)=2411.

Is this solution Helpfull? Yes (4) | No   

DIVYA V M  (10 years ago)

Sum Cant be more than double the value of terms that is 1230. sum cnt be more than 2460 since none of the term is more than 2.

Is this solution Helpfull? Yes (1) | No   

PRATIM JANA  (8 years ago)

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 (1) | No   

lavanya  (10 years ago)

Write the series as
1
2 1
2 2 1
2 2 2 1.......add each line of terms then...the series would be...
1 3 5 7 ........upto 492 terms(first four numbers means sum of 10 numbers)
last term=a+(n-1)d=1+(492-1)2=983
sum of n terms =(n(a+l))/2=242664

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lavanya  (10 years ago)

1............(1)
2 1..........(3)
2 2 1.........(5)
2 2 2 1.......(7)
that means 10 terms sum is (1+3+5+7)..so next 26 terms(9+11+13+15)....
so terms series would be...
10 26 42 58 74 90 106 122 138 154 170 186(sum is 1176)...next one is 202..but if e add them it will be 1378..we need only 1230...still 54 terms are to be added.
1+3+5+7+9+11...........48 terms(12*4)=sum is 2304
in that 54 terms(48*2+1=97 i.e., total 49 terms still 5 2's are needed )
so total would be 2304+97+(5*2)=2411
ans 2411

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deep chandra bhatt  (10 years ago)

Rakesh bhai solution repeat n=48 kaise aaya

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velampati munipriya  (8 years ago)

can u pls tell me ..hw u taken n=48

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