There are 25 necklaces such that the first necklace contains

Question

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

A. 2690
B. 3025
C. 3380
D. 2392
E. 3762

Show Answer

Official Answer

B

A. 2690
B. 3025
C. 3380
D. 2392
E. 3762

Bunuel · Accepted Answer

There are 25 necklaces such that the first necklace contains 5 beads, the second contains 7 beads and, in general, the ith necklace contains i beads more than the number of beads in (i-1)th necklace. What is the total number of beads in all 25 necklaces

A. 2690
B. 3025
C. 3380
D. 2392
E. 3762

Show Answer

Official Answer

B

A. 2690
B. 3025
C. 3380
D. 2392
E. 3762

I think above solution is not correct. I also think it's not GMAT question as it requires way more than 2 minutes to solve this.

We have sequence:  5, 7, 10, 14, 19, 25...   (25 terms)  and we want to determine the sum.

Let's subtract  4  from each term it will total  4*25=100 , and we''l get:

100+1+(3+6)+(10+15)+(21+28)+...  After 100 we have the same 25 terms.

100+1^2+3^2+5^2+7^2...  after  100  we have the sum of the squares of the first n odd numbers. As there are  25  terms after  100  we'll have  n=13  squares.

The sum of the squares of the first  n  odd numbers= \frac{n*(2n-1)(2n+1)}{3}=\frac{13*25*27}{3}=2925 .

100+2925=3025

Answer: B (3025)

ezinis · Answer

let me know your comments:
a1 = 5

a1 = a1
a2 = a1 + 2
a3 = a1 + 2 + 3
.
.
.
a24 = a1 + 2 + 3 + ...... + 24
a25 = a1 + 2 + 3 + ...... + 24 + 25

Total = 25*5 + 24*2 + 23*3 + ....3*23+ 2*24 + 25
 = 26*5 + definae Integral of x(26-x)dx from 2 to 24
 = 2961 ( it took me 5 min to do by hand, i checked with Mathcad for the integral part)

sriharimurthy · Answer

Im trying to think up a faster way to solve this but nothing seems to be striking

Im sure there must be one else this problem is way to time consuming to be on the gmat...

ezinis · Answer

I was correct till the definite integral part as I make the curve smooth ...

Anyway, I recalculate the sum above and it it is equal to 3025. Bravo Bunuel. +1 from me.

ezinis · Answer

anyone could come up with a faster way to calculate this other than muscle??:
24*2 + 23*3 + ....3*23+ 2*24 = 2*(24*2 24*2 + 23*3 + ....14*12) + 13*13

jade3 · Answer

The answer is B. I would be happy to skip this problem on the GMAT, if it comes. It has taken me ages to solve this

Any way Please comment whether this solution is faster

We have a sequence of 5, 7, 10, 14, 19…. = (4+1), (4+3), (4+6)…….
The sum of the sequence is = (25*4)+1+3+6+….25(25+1)/2 (Generic term is x(x+1)/2)

1+3+6+….25(25+1)/2 = Sum of first 25 triangular number (The formula of sum of n triangular number is ½  +Sum ], when n=25; ½  +Sum ] =2925)

25*4+ 2925 =3025

amit2k9 · Answer

5+ (5+2) + (5+5) + (5+9) .....
25*5 + (2+5+9+14+...)

In (2+5+9+14+ ...) the differences are in AP with a= 3, and common difference = 1

for 2nd term difference = 3
for 3rd term difference = 4
................................
for 25th term difference = 26

adding 3,4,5,,26 =  *24(number of terms) = 348

last difference = 348. Last term = 350

I will try to figure out more on this now.

sagarsir · Answer

We are given T(n) = T(n-1) + n ----------1
and T1 = 5 ------- 2

We apply Summation (S) on and add constant to T(n) that we obtain containing only n

S  = S  S

S  - S  = S  ----- difference of Sum of all and and the sum of all but last term is T(n)

T(n) = S

Thus T(n) = n(n+1)/2 + k

from 2, we get T(1) = K = 4

Thus T(n) = (N^2 + N) / 2 + 4

Apply summation on T(n) to get the required sum

S(n) = 1/2[ n*(n+1)*(2n+1)/6 + n(n+1)/2) + 4n

Substitute N = 25 Answer = 3025.

KUDOS if u like!!!!

fdecazaux · Answer

Let us name N the total number of beads on the 25 necklaces and ni the number of beads on the ith necklace.

N= n1 + n2 + + n25
n1 = 5
n2 = 5 + 2
n3 = 5 + 2 + 3
n4 = 5 + 2 + 3 + 4

n25 = 5 + 2 + 3 + 4 + 25

N = 5 + (5+2) + (5+2+3) + (5+2+3+4) + + (5+2+3+4+ 25)
By rearranging the terms, we find
N = (25*5) + (24*2) + (23*3) + (22*4) + (1*25)
N = 125 + Sum ((25-k)*(k+1)) from k = 1 to 24 
N = 125 + Sum (24k-(k^2)+25) from k = 1 to 24
By developing and applying the formulas Sum (k^2) = 1/3 N*(N+0,5)*(N+1) (for k = 1 to N)
We find N = 3025

Perhaps · Answer

in the above solution i understood each n every step...
but i have only one concern .....from where it wud enter in our minds during the exam that v got to subtract '4' from each term to get the sequence....
is there any way to observe this thing....i know it wud come only by practice but still i wanted to know that how it wud strike our minds to subtract 4 from each term.....phewwww!!!!!!!!

jlgdr · Answer

Here's another way of solving quickly

We have a sequence of 5,7,10,14,19,25 etc

Note that we have pairs O-O-E-E-O-O etc... We are going to have 12 pairs and one left, hence the sum of the terms will be ODD

Only ODD in the answer choices is B

jlgdr · Answer

Bunuel, how did you think about reducing 4 in each number to find perfect squares? What's the link between perfect squares and that sequence?

Thanks
Cheers
J

sagnik242 · Answer

How did you know after 7 to go on to 10 and so forth? I know you probably used the i more beads info but can you show the plugging in the #'s to get 10 if this is what you did?

sagnik242 · Answer

Actually Bunuel,

I am also confused on how exactly you arrived at 100 + 1 + (3+6) + (10+15) + (21+28)+... I know you said subtract 4 from each term and then 4 x 25 = 100 but I don't see where all these terms are coming from.

Also I don't get the step after which is 100 + 1^2 + 3^2 etc... where is the 1^2 , 3^2 coming from?

Hope you have a great 4th of July!

sagnik2422 · Answer

HOW DID YOU KNOW THE NUMBERS AFTER 7? CAN YOU PLEASE SHOW HOW YOU PLUGGED IN AND FIGURED OUT? Thanks

Bunuel · Answer

1st necklace = 5 beads;
2nd necklace = 5 + 2 = 7 beads;
3rd necklace = 7 + 3 = 10 beads;
4th necklace = 10 + 4 = 14 beads;
...

Bunuel · Answer

100 + (5 - 4) + (7 - 4) + (10 - 4) + (14 - 4) + (19 - 4) + ...
100 + 1 + 3 + 6 + 10 + 15 +...
100 + 1 + (3 + 6) + (10 + 15) + ...
100 + 1 + 9 + 25 + ...
100 + 1^2 + 3^2 + 5^2 + ...

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

There are 25 necklaces such that the first necklace contains

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

There are 25 necklaces such that the first necklace contains

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards