Latest Elitmus Aptitude Question SOLUTION: 1,3,6,10,15,21,28........................so on up to 300 terms then find the sum

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(#M40038313) ELITMUS Numerical Ability question Aptitude Keep an EYE

1,3,6,10,15,21,28........................so on up to 300 terms then find the sum

Asked In Elitmus T.VIGNESWARA REDDY (9 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

Saraswathy  (9 years ago)

1 ,3, 6, 10 is of the form n*(n+1)/2
n = 300
nth term = 300*(301)/2 = 45150
Then sum of these terms 1+3+6+10+15+......45150 ll be
Sum of n*(n+1)/2 terms
Sum of n*(n+1)/2 terms = 1/2* sum of [n^2 + n] terms
Sum of n^2 terms => n(n+1)(2n+1)/6
sum of n terms => n(n+1)/2
So Sum ll be 1/2 * [n(n+1)(2n+1)/6 + n(n+1)/2]
n(n+1)(2n+4)/12
n = 300
So 300(301)(604)/12 = 4545100

Ans : 4545100

Is this solution Helpfull? Yes (38) | No   

Mohit Anand  (9 years ago)

each term in the series follows the equation-n(n+1)/2, where n is the position
therefore for the sum of 300 terms of the series , we will apply summation over n(n+1)/2,
hence it will expand to-- ((n(n+1)(2n+1))/6)+(n(n+1)/2)
which will compute to 300*301*604/12
=4545100

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raushan kumar  (9 years ago)

ans==4545100

Is this solution Helpfull? Yes (3) | No (3)  

mareeswari  (9 years ago)

sn=n(a+l)/2.....
n=300,a=1,l=300
sn=300(301)/2
ans=45150

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VIJAY SINGH  (8 years ago)

=1+3+6+10+15+21+28+36+.............................up to 300 terms
=4+16+36+64+..................................up to 150 terms
=2^2+4^2+6^2+8^2+...........................up to 150 terms
=2^2*(1^2+2^2+3^2+4^2...........................up to 150 terms)
=4*(n*(n+1)*(2*n+1)/6)
put n=150
=2*(150*151*301)/3
=4545100

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Goldi kumari  (9 years ago)

n=300 and a=1 and d=1 so now we no the formula of ap series n/2[2a+(n-1)d]
300/2[2*1+(300-1)*1]
45150

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ritwik misra  (9 years ago)

can someone tell me wat will be d diagram..i am got getting the question properly

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sivakumarvalavala  (9 years ago)

n=300,a=1,d=1,(the diffrence between successive terms in A.P..1,3-1,6-3,10-6,15-15....) we know the formula.sum of the nth terms in A.P=n/2(a+l)
=300/2(1+300)=45,150

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Kumar Anmol   (8 years ago)

term are in the of n*(n+1)/2
sum of n^2 is n*(n+1)*(2*n+1)/12
sum of n term is n*(n+1)/2
so, sn=(n*(n+1)*(2*n+1)/12+n*(n+1)/2)/2
so ans 4545100

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