Latest Elitmus Aptitude Question SOLUTION: The sum of how many terms of the series 6 + 12 + 18 + 24 + … is 1800 ?A)16B)24C)20D)18E)22

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The sum of how many terms of the series 6 + 12 + 18 + 24 + … is 1800 ?A)16B)24C)20D)18E)22

Asked In Elitmus divyasree (8 years ago)

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Vikas William  (8 years ago)

ANS->B. 24
1800=N/2[2A+(N-1)D]

Is this solution Helpfull? Yes (18) | No (1)  

Mrigendra Pratap Singh  (7 years ago)

Let the number of terms be N.
Then, according to the formula,
=> 1800 = N/2[2A+(N-1)D] , where A = First-Term, D = Common-Difference.
=> 3600 = N[12 + (N-1)6]
=> 3600 =N[6 + 6N]
=> 6N^2 + 6N - 3600 = 0.
=> 6N^2 + 150N - 144n - 3600 = 0
=> 6N(N + 25)-144(N+25)
=> N = -25, 24
Hence, the number of terms is 24.

Is this solution Helpfull? Yes (9) | No (2)  

Rahul Kumar TIwari  (8 years ago)

if you take 6 common among all number then it can be easily solve using n(n+1)/2.

Is this solution Helpfull? Yes (3) | No (8)  

vinod kumar korikana  (7 years ago)

sum of the numbers to a number n is n(n+1)/2
so 6*(n(n+1)/2)=1800
n(n+1)/2=300
on solving ,n=24

Is this solution Helpfull? Yes (2) | No   

vijay  (8 years ago)

300 terms of series gives 1800.

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Shalinee Shankar  (7 years ago)

we will solve it by sum of n terms....s=n/2[2a +(n-1)d]...........so ans.will be 24

Is this solution Helpfull? Yes (1) | No   

Naveen babu  (7 years ago)

=6*1+6*2+6*3+6*4+..........
=6(1+2+3+4+.............up to n terms)
= 6( sum of first n terms )
=> 6(n(n+1)/2) = 1800
by solving above we get n= 20

Is this solution Helpfull? Yes (1) | No (3)  

KUMAR HARSH  (7 years ago)

a=6;d=6;s=1800
1800=n/2[2a+(n-1)d]
1800=n/2[12+6n-6]
1800=n/2[6+6n]
1800=n[3+3n]
600=n(1+n)
600=24*25
n=24

Is this solution Helpfull? Yes (1) | No   

Ajeet Kumar  (8 years ago)

last term=a+(n-1)d
1800=6+(n-1)6
300=1+(n-1)
n=300

Is this solution Helpfull? Yes | No (13)  

PENDYALA SASI SEKHARA VAIBHAV  (7 years ago)

24. Since, n2+n-600=0, n=24

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