Latest TCS Aptitude Question SOLUTION: The sum of first 40 terms 1,6, 7,12, 13,20 …. ?

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(#M40039376) TCS Numerical Ability question Keep an EYE

The sum of first 40 terms 1,6, 7,12, 13,20 …. ?

Asked In TCS bhagwandas kushwaha (9 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

Sushmita saxena  (9 years ago)

[1, 7, 13.......20 terms] + [6, 12, 20......20 terms]
[1 + (20-1)*6 ] + [this forms (n+1) (n+2)]
117 + [ n^2 + 3n + 2]
117 + 2870 + 630 + 2
= 3715

Is this solution Helpfull? Yes (6) | No (7)  

Shashi Shekhar  (9 years ago)

Observing the sequence 1, 6, 7, 13, 20, 33, and so on,
It can be re-written as 1, (1+5), (2+5), (3+10), (5+15), (8+25), and so on.

The sequence thus decomposes into sum of two standard Fibonacci sequences,
1, 1, 2, 3, 5, 8, 13 up to 40 terms and 5*(1, 1, 2, 3, 5, 8, 13 up to 39 terms.

So the sum is given as : Sum of 40 terms of the standard Fibonacci series + 5*Sum of 39 terms of the standard Fibonacci series. Which is equivalent to 6*Sum of 39 terms in the standard Fibonacci series + the 40th term. And since sum of n terms of a fibonacci series is given as F(n+2)-1,

The sum should be 6*{F(41)-1}+F(40) = F(40) + 6*F(41) - 6.

Check the values!

Is this solution Helpfull? Yes (2) | No (5)  

GAUTAM KUMAR  (9 years ago)

1 add & 1 delete on series
ans 2420

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

mounika  (9 years ago)

hw can we find F(41),F(40) shekhar sir.

Is this solution Helpfull? Yes (1) | No   

B.Sriteja  (9 years ago)

you missed the 12 in the series..

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venkata satya kumar  (9 years ago)

by using the formulas an=a1+(n-1)d and
sn=n(a1+an)/2 we can get the solution

Is this solution Helpfull? Yes | No (2)  

upasana gupta  (9 years ago)

please explain solution for this more clearly....

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Arun kumar  (9 years ago)

Observing the sequence 1, 6, 7, 13, 20, 33, and so on,
It can be re-written as 1, (1+5), (2+5), (3+10), (5+15), (8+25), and so on.

The sequence thus decomposes into sum of two standard Fibonacci sequences,
1, 1, 2, 3, 5, 8, 13 up to 40 terms and 5*(1, 1, 2, 3, 5, 8, 13 up to 39 terms.

So the sum is given as : Sum of 40 terms of the standard Fibonacci series + 5*Sum of 39 terms of the standard Fibonacci series. Which is equivalent to 6*Sum of 39 terms in the standard Fibonacci series + the 40th term. And since sum of n terms of a fibonacci series is given as F(n+2)-1,

The sum should be 6*{F(41)-1}+F(40) = F(40) + 6*F(41) - 6.

Check the values!

Is this solution Helpfull? Yes | No (1)  

aman sharma  (9 years ago)

ans 5443
divide into two parts
20 terms of 1,7,13 an a.p use formula
an for other on 6=3*2
12=3*4
and so one then calculate values and add....seens lenghti but takes only 2 mins

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