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

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

Asked In TCS Suman Saurabh (10 years ago)

Unsolved Read Solution (18)

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Read Solution (18) :

prabhat kumar  (10 years ago)

in this series two series ,
1st is 1,7,13...115
this series sum=n/2[2a+(n-1)d]=1160.
and second series is
6,12,20,30,42,54,....432
sum=3390.
so,total sum=3390+1160=4550

Is this solution Helpfull? Yes (7) | No (14)  

Paras Jain  (9 years ago)

The given series consists of 2 APs...
AP1: 1,7,13... use sum formula => sum = n/2[2a+(n-1)d] , here a=1 n=20 d=1;
so sum = 1160.

AP2: 6,12,20...
or 2*3, 3*4, 4*5.....
****IMPORTANT****
this is AP is further broken down into 2 APs

AP1 : 2,3,4,5..
AP2 : 3,4,5,6..

now to find the sum..we will use summation (sigma)....

sigma(n=1 to 20) [a1+(n-1)d]*[a2+(n-1)d] ,,,,,,,here a1=2 and a2=3

so it becomes summation of sigma (n=1 to 20) n^2 + 3n + 2...

this can be solved using summation n^2 = n(n+1)(2n+1)/6
summation 3n = 3n(n+1)/2
summation 2 = 2n


put n=20 in all...
sum = 3540..

Ans. is 1160 + 3540 = 4700

Hope it helps..!!

Is this solution Helpfull? Yes (4) | No (1)  

medapati nagendra reddy  (10 years ago)

sorry my above approch is wrong...

Is this solution Helpfull? Yes (3) | No   

Suvojyoti Sinha Ray  (10 years ago)

Answer is 4700.. Verified!

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

Pratyusha Bhadra  (10 years ago)

6,12,20....it is nt an AP.so hw to get the sum of 20 terms?plz explain

Is this solution Helpfull? Yes (2) | No   

VISHAKHA BHATTACHARJEE  (10 years ago)

the 2 series are:-
1,7,13,..
series is in ap so sum is = n/2[2a+(n-1)d]=1160 (n=20 and a=1)
the second series is:-
6,12,20,30,...
the difference of the above series is in ap ie the difference is :- 6,8,10,..
so the last difference of terms will be= a+(n-1)d=42 (a=6 and n=19 and d=2)
so the second series is:- 6,12,20,30,....460
sum of the second series=3510
total sum=3510+1160=4670

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

manish kumar  (10 years ago)

sum of first series=1160
for second 6,12,20....462=3540

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Naba Paul  (10 years ago)

19...7-1=6,13-7=6,19-13=6

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madhurima gorai  (10 years ago)

1095814995

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medapati nagendra reddy  (10 years ago)

dividing the above sereis into two groups
1,7,13,.....139(total 20 terms)
6,12,18,....126(total 20 terms)
using formula sum of terms =n/2(2a+(n-1)d)
for first sereis n=139,a=1,d=6 substituting in above formula we get 57685
again for second sereis n=126,a=6,d=6 sum=48006
sum of two sereis=57685+48006=105691 is the answer

Is this solution Helpfull? Yes | No (11)  

chandan kumar  (10 years ago)

7/10
in above i explain but mistake in total term

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Ashu Pal  (10 years ago)

we have two series of AP
first--- 1 7 13......
2nd-- 6 12 20......
calculate individual sum of each
first-- Sum=560
2nd--- 660
Total = 560+660=1120
ans is 1120

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pre  (10 years ago)

m not getting it..pls explain?

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virat pal  (10 years ago)

4700
sum of ap + sum of 6,12,20

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sandeep gantait  (10 years ago)

in the given question there are two different series:
1,7,13....
last term/20th term if the series: 1+(20-1)x6=115
so the series is 1,7,13....115
sum of the series is 20/2(1+115)=1160

now consider the second series:6,12,20....
we could break it in to two series:
1x6,2x6+0,3x6+2,4x6+4,5x6+6.....
which could be break down in to
6,12,18....120(up to 20th term)
Sum:1260
and
2,4,6....36(up to 18th term)
sum:162
TOTAL SUM:2582(adding all 3 sum)

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mudy  (10 years ago)

ans=4700..its correct verifiedand

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Chirag Ranjan Lahiri  (9 years ago)

4700
sum of ap + sum of 6,12,20

Is this solution Helpfull? Yes | No (1)  

tripti grover  (9 years ago)

how dis you find the sum of second series?? i.e 6,12,20,,.....?? please elaborate..

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