Latest TCS Aptitude Question SOLUTION: sum of 52 terms of the series 1,6,7,13..

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(#M40029584) TCS question Keep an EYE

sum of 52 terms of the series 1,6,7,13..

Asked In TCS Paramita Saha (10 years ago)

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Jeet Das  (10 years ago)

1+6+7+13+20+.....up to 52 terms
=1+6+(6+1)+(6+6+1)+(6+6+6+1+1)+....+[(6+6+...upto 50 terms)+(1+1+....+up to 49 terms)]
A.P series from 3rd term....6[1+2+...+50]=6*25*51
A.P series from 4th term.... [1+2+....+49]=49*25
and the rest 1+6+1=8
so, total of above series=6*25*51+49*25+8=8883

Is this solution Helpfull? Yes (13) | No (14)  

Shirshendu Ghosh  (10 years ago)

Consist of two A.P. series
1,7,13.. and 6,13,20 with common difference of 6 and 7 respectively.
So, sum of 1st series 1974 as last term i.e., 26th term is 151
sum of 2nd series 2418 as last term 187.
so sum of 52 terms 2418+1974=4394

Is this solution Helpfull? Yes (4) | No (21)  

shubham kumar  (10 years ago)

@shirsedu ghosh your logic is not right.

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Sriparna Boote  (10 years ago)

Its a fibonacci series

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Pratyusha Bhadra  (10 years ago)

is there any formula to calculate sum of Fibonacci series?

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payaljain  (10 years ago)

find nth term which is summ of n-1th and n-2 term
Tn=n-1+n-2=2n-3
use summation and find ans is 2652

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Shirshendu Ghosh  (10 years ago)

@subham kumar i admit. Can you help me out?

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jayati nandy  (10 years ago)

can sum1 pls xplain ds??

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virat pal  (10 years ago)

5004
1+7+...

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DIVYA KUKRETI  (10 years ago)

ans: 4403
It can be broken into to A.P series
A.P 1= 1+7+13....
A.P 2= 6+13+...
COMMON DIFFERENCE= (7-1)=6 in A.P 1
=(13-6)=7 in A.P 2
no of terms in A.P 1 AND 2=(52/2)=26
SO, By formula s=n/2(2a+(n-1)d)
we get-
for A.P 1=1976
for A.P 2=2431
adding both=4403

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