Latest TCS Aptitude Question SOLUTION: Find the 100th term of the following series - 2,4,8,14,22,32...

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(#M40011423) TCS question Keep an EYE

Find the 100th term of the following series - 2,4,8,14,22,32...

Asked In TCS saurabh verma (11 years ago)

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Rupesh kashyap  (11 years ago)

t(n+1)=n^2+n+2
t(0+1)=0+0+2=2
t(1+1)=1^2+1+2
.
.
.
.
t(99+1)=99^2+99+2=9902

Is this solution Helpfull? Yes (38) | No (2)  

NILOY SUTRADHAR  (11 years ago)

2nd term -1st term=1*2
3rd term -2nd term=2*2
4th term -3rd term=3*2
.....................
.....................
nth term -(n-1)th term=(n-1)*2
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sum:nth term -1st term=[1+2+3....(n-1)]*2(as rests are cancelld)
so 100th term=(100*99/2)*2+2(as sum=n(n-1)/2,n=100,1st term =2)
=9902ans

Is this solution Helpfull? Yes (5) | No (1)  

karthikeyank  (9 years ago)

ANS IS : 9902
a(n) = 2 + n(n - 1)
a(100)=2+100(100-1)
=9902

Is this solution Helpfull? Yes (3) | No   

nirvan gupta  (11 years ago)

t(n)=t(n-1)+2n ;n>0
t(0)=2
t(1)-t(0)=2
t(2)-t(1)=4
.
.
.
t(100)-t(99)=200
summing up we get t(100)=(2*100*101/2)+t(0)=10102

Is this solution Helpfull? Yes (2) | No (7)  

SAKTHI SARAVANAN.V  (11 years ago)

2 4 8 14 22 32
diff 2 4 6 8 10
again diff 2 2 2 2
now apply formula tn=a+(n-1)d
t100=2+(99)2
t100=200

Is this solution Helpfull? Yes (2) | No (6)  

Rupesh kashyap  (11 years ago)

@sumit gupta----
actually the t(n) comes from the given logi....its not aaplicable for all questions..

s(n)=3+6+12+...+t(n)------eq 1
s(n)= 3+6+12+...+t(n)+t(n+1)------eq2
eq1-eq2
t(n+1)=3+(3*n)/2+(3*n*n)/2
this is d correct way....enjoy

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Sumit Gupta  (11 years ago)

@rupesh
Find the 100th term of the following series - 3,6,12,21,33,48...
now what will be the logic

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niki  (11 years ago)

can we solve the series 3,6,12,21,33.... using the above method i.e.
t(n+1) = n^2+ n+2?

Is this solution Helpfull? Yes | No (1)  

c.r.k sagar  (11 years ago)

@sumit...u rfinding the nth term for series of common differences and not for the given series!!

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shyamlesh dash  (11 years ago)

.Now, How can you find any term like say 22 if you're given 2 only and this difference. Just the some of the differences plus 2.=> 2+ (2+4+6+8) = 22. Hence, 100th term = Sum of AP (2,4,6,8...) till 99 terms + 2=> [ 99*(2+98*2)/2 ] + 2 = 99*99 + 2 = 9801 + 2 = 9803.

Is this solution Helpfull? Yes | No (7)  

shivesh  (11 years ago)

here nth term =2*n-1+n-1th term
so 100 term=2*99+99th term
99th term=2*98+98th term and so on
s0
2(99+98.....1)+2=9704

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