Latest TCS Aptitude Question SOLUTION: the sum of first 40 terms 1,6,7,13,20.........??????? give slon

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(#M40028533) TCS question Keep an EYE

the sum of first 40 terms 1,6,7,13,20.........???????
give slon

Asked In TCS TCS 2014 (10 years ago)

Unsolved Read Solution (6)

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raja babu  (10 years ago)

sum for this fibonacci series is S(n)= S(n+2)-S(2)
so for finding the sum upto 40th term we have to find 42nd term
f(42)=1095815001 (using C program :p)
so S(40)= S(42)-S(2)
S(40)=1095815001-6
so ans is S(40)= 1095814995

Is this solution Helpfull? Yes (5) | No (7)  

chaitanya  (10 years ago)

sry solution is not completed

Sn=n(a1+an)/2
=40(236)/2
=40x118
=4720

Is this solution Helpfull? Yes (4) | No (12)  

raja babu  (10 years ago)

sum for this fibonacci series is S(n)= f(n+2)-f(2)(u can check by taking smaller value of n)
so for finding the sum upto 40th term we have to find 42nd term
f(42)=1095815001 (using C program :p)
so S(40)= f(42)-f(2)
S(40)=1095815001-6
so ans is S(40)= 1095814995

Is this solution Helpfull? Yes (3) | No (2)  

Debolina Chakraborty  (10 years ago)

as 1+6+7 is sum upto 3rd tem n it is 5th term -6, similarly
as 1+6+7+13+20+33+53 is sum upto 7th term and is equal to 9th term 139-6=133
hence sum upto 52nd term is 54th term-6
using golden ratio we can solve this, bt to be more accurate we calculate a few more terms. 1,6,7,13,20,33,53,86,139,225,364,589,953,1542....
thus 54th term= 1542*(1.6180339)^40
hence sum upto 52nd term is 352849113676-6=352849113670

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

palak  (10 years ago)

see,
1-------sum upto 1st term
7-------sum upto 2nd term
14-------sum upto 3rd term
27-------sum upto 4th term
47-------sum upto 5th term
80-------sum upto 6th term
133-------sum upto 7tht term
219-------sum upto 8th term

see Sum uptoNth term=(N+2)th term-6 (for this series.)

so sum upto 52th term=54th term-6

1+6+7+13+20+33+53=sum upto 7th term= 9th term-6=139-6
NOW USING "GOLDEN RATIO" WE CAN FIND THAT

1,6,7,13,...........................,139,225,364,589,953,1542(we have taken here 14TH TERM for accuracy)

so 54th=1542(14th term)*(14th term/13th term)^40

as we considered here 14th terms so the power is (54-14)=40
so we get,
1542*(1.6180339)^40=352849113676

so sum upto 52th=54th - 6
=352849113676 - 6
=352849113670

Is this solution Helpfull? Yes (1) | No   

chaitanya  (10 years ago)

think so the question is mistaken it must be like 1,7,13,20........
then
First find the 40th term:

a40 = a1+(n-1)d

given a1=1

d=a2-a1

=7-1

=6
=> a40= 1+39(6)
= 1+234
= 235
Then find the sum:

Sn=(a1+an)/2
=(1+235)/2
= 236/2
= 118

Is this solution Helpfull? Yes | No (28)  

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