Latest Elitmus Aptitude Question SOLUTION: Q. 1,2,1,2,2,1,2,2,2,1,2,2,2,2... then sum of 1230 terms...?

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(#M40032633) ELITMUS Numerical Ability question sum of the terms Keep an EYE

Q. 1,2,1,2,2,1,2,2,2,1,2,2,2,2... then sum of 1230 terms...?

Asked In Elitmus rahul avinash (10 years ago)

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KAMESH TYAGI  (9 years ago)

(1,2),(1,2,2),(1,2,2,2),(1,2,2,2,2)........
(2+1),(1+2+2),(1+2+2+2)..........
3,5,7,9........
Here a=3, d=2, n=1230
last term=a+(n-1)d=3+(1230-1)2=2461
sum=n/2[2a+(n-1)d)=1230/2[(2*3)+(1230-1)*2]=1515360 ans

Is this solution Helpfull? Yes (6) | No (5)  

sweta singh  (10 years ago)

(12) (122)(1222)..............
let there be n series of(12222.....)have the 1230 terms
2+3+4+5+...........n=1230
n(n+1)=2462
we have..49*50+12=2462(so n=49)
so series will be
(12)(122).........................(1222...49 times)(12222....11 times)
now summing up these terms
2(1+2+3+.......+48+11)+50(1)
=2444

Is this solution Helpfull? Yes (1) | No (6)  

MR. X  (9 years ago)

(1,2),(1,2,2),(1,2,2,2),(1,2,2,2,2)........
(3) , (5) , (7) , (9) .........
(2) , (3) , (4) , (5) ........ now these series sum equals 1230 according to question
sum=n/2[2a+(n-1)d)=>n=48.12 (sum=1230.a=2,d=1)
find sum upto 48 term by formula ( sum=n/2[2a+(n-1)d])=>1223;
more 7 term are required (ie 1,2,2,2,2,2,2(sum=13))

now therefore final result will be sum of the 2nd series(3,5,7,9...... upto 48th term)+13
=2400+13
=2413 ans

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Sourabh Prasad  (8 years ago)

Divide the series in two parts.
1,1,1,1,1,.......
2 , 2 2 , 2 2 2, 2 2 2 2, ...................

i.e, sum is like
1 2 3 4 5 6...... of 1st series
2 4 6 8 10 .......... of 2nd series

so in general it is like
n
n(n+1)
i.e,
n+n(n+1)
=n²+2n

We have find sum of 1230 term
Just put n=1230
sum= 1515360

Is this solution Helpfull? Yes (1) | No   

Madhu  (6 years ago)

Solution:

Group the no. As follows:

X= {1, 2+1, 2+2+1, 2+2+2+1, 2+2+2+2+1,…}

Now it gives you A.P. as follows:

Y= {1, 3, 5, 7, 9,…}

Number of terms contained in each Y is:

Z= {1, 2, 3, 4, 5…}

Given the sum to be found ulti 1230 terms but 1230 doesn't give you the exact number if you solve it by sum of n natural no.s formulae. So break 1230 into 1225+5.

Now find the value of n given the sum is 1225.

n(n+1)/2 = 1225

This gives n=49

Now use the sum of A.P. formulae in Y:

Sum= n[2a+(n-1)d]/2

Where n=49, a=1, d=2 we get:

Sum = 2401

Now this is the result for 1225 no.s

To get the result of 1230 no.s we need to add the next 5 numbers in the sequence (which starts with 2). So the next 5 terms are 2,2,2,2,2.

Therefore =2*5 = 10

Final result is 2401+10 = 2411

Is this solution Helpfull? Yes (1) | No   

Prasanth  (10 years ago)

ans: 2353

12
122
1222
12222
......
consider a square(with 2460 elements) sliced half diagonally. Sqrt(2460)= 49
==>49X49(outer square with 1)and 48X48(inner square with 2)
==>48^2=2304 elements, half diagonal ==> 2304/2= 1152 with each value of 2==> 1152*2=2304
1152(2's)+49(1's(only one column))
==>2353

Is this solution Helpfull? Yes | No (3)  

niharika pandey  (9 years ago)

In starting 10 terms 1212212221,
the number of 1s is 4
and the number of 2s is 6
so 1s are 40% and 2s are 60% in the series.............
so in 1230 there are 492 1s will be there
and 738 2s....................
so 2*732 + 492 = 1968..............

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