Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 87)
87.
The sum of how many terms of the series 6 + 12 + 18 + 24 + ... is 1800 ?
Answer: Option
Explanation:
This is an A.P. in which a = 6, d = 6 and Sn = 1800
| Then, | n | [2a + (n - 1)d] = 1800 |
| 2 |
 |
n | [2 x 6 + (n - 1) x 6] = 1800 |
| 2 |
3n (n + 1) = 1800
n(n + 1) = 600
n2 + n - 600 = 0
n2 + 25n - 24n - 600 = 0
n(n + 25) - 24(n + 25) = 0
(n + 25)(n - 24) = 0
n = 24
Number of terms = 24.
Discussion:
15 comments Page 1 of 2.
Likhitha said:
1 decade ago
Please tell me why we have taken this formula, i.e n/2[2a+(n-1)]d.
In the previous problem of (2+4+6+...30) we have taken the formula as [a+(n-1)d]. So tell me which formula must be used for which problem.
In the previous problem of (2+4+6+...30) we have taken the formula as [a+(n-1)d]. So tell me which formula must be used for which problem.
Tanuja said:
1 decade ago
In the previous problem we know the last term. If we know the last term we can use the formula [a+(n-1)d] = L(last term) then we can find the number of terms n and we can calculate sum of terms using n/2(a+l).
But in above case we know the sum of terms is 1800 so using sum of n terms formula n/2[2a+(n-1)d] we will calculate the number of terms.
But in above case we know the sum of terms is 1800 so using sum of n terms formula n/2[2a+(n-1)d] we will calculate the number of terms.
Aparna said:
1 decade ago
Please explain me how could it become as n (n+1) = 600.
From this step to last step I did not understood please anyone help me to know.
From this step to last step I did not understood please anyone help me to know.
Neeraj Patel said:
1 decade ago
Please explain me step 3rd.
How you get 3n (n+1) = 1800?
How you get 3n (n+1) = 1800?
Albin Antony said:
9 years ago
Another method:
6+12+...... = 1800,
6(1 + 2 + 3 + ....) = 1800,
1+2+3.... = 300,
So, n(n+1)/2 = 300,
n(n+1) = 600.
6+12+...... = 1800,
6(1 + 2 + 3 + ....) = 1800,
1+2+3.... = 300,
So, n(n+1)/2 = 300,
n(n+1) = 600.
Suyog said:
9 years ago
Nice move @Albin.
(1)
Vetri said:
8 years ago
Please explain me 3rd step.
How you get 3n(n+1)=1800?
How you get 3n(n+1)=1800?
(1)
Lahari said:
7 years ago
Please explain me the 3rd step.
(1)
Senbagam said:
7 years ago
How you get 3n(n+1) = 1800?
Solution is;
n/2 [2 x 6 + (n - 1) x 6] = 1800,
n/2 [12 + (6n-6)] = 1800,
12n/2 + 6n2/2 - 6n/2=1800,
12n+6n2-6n/2=1800,
6(2n+1n2+1n)/2=1800,
6(1n2+1n)/2=1800,
3(1n2+1n)=1800,
3n(n+1)=1800.
Solution is;
n/2 [2 x 6 + (n - 1) x 6] = 1800,
n/2 [12 + (6n-6)] = 1800,
12n/2 + 6n2/2 - 6n/2=1800,
12n+6n2-6n/2=1800,
6(2n+1n2+1n)/2=1800,
6(1n2+1n)/2=1800,
3(1n2+1n)=1800,
3n(n+1)=1800.
(3)
Ruben said:
7 years ago
(6+6x)x/2 = 1800.
(6+6x)x = 3600,
6x^2+6x-3600 = 0,
6(x^2+x-600) = 0,
x^2+x-600 = 0,
x^2+25x-24x-600 = 0,
x(x+25)-24(x+25) = 0,
(x-24) or (x+25).
x=24 or x= -25.
So ans n=24.
(6+6x)x = 3600,
6x^2+6x-3600 = 0,
6(x^2+x-600) = 0,
x^2+x-600 = 0,
x^2+25x-24x-600 = 0,
x(x+25)-24(x+25) = 0,
(x-24) or (x+25).
x=24 or x= -25.
So ans n=24.
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