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

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.
8 years agoHelpfull: Yes(9) No(2)
if you take 6 common among all number then it can be easily solve using n(n+1)/2.
8 years agoHelpfull: Yes(3) No(8)
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
8 years agoHelpfull: Yes(2) No(0)
300 terms of series gives 1800.
8 years agoHelpfull: Yes(1) No(11)
we will solve it by sum of n terms....s=n/2[2a +(n-1)d]...........so ans.will be 24
8 years agoHelpfull: Yes(1) No(0)
=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
8 years agoHelpfull: Yes(1) No(3)
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
8 years agoHelpfull: Yes(1) No(0)
last term=a+(n-1)d
1800=6+(n-1)6
300=1+(n-1)
n=300
8 years agoHelpfull: Yes(0) No(13)
24. Since, n2+n-600=0, n=24
7 years agoHelpfull: Yes(0) No(0)
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7 years agoHelpfull: Yes(0) No(0)

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