Find the 51st term of the series.

Question

A particular series is given by :

A1 = {1}
A2 = {2,3,4}
A3 = {5,6,7,8,9}
A4 = {10,11,12,13,14,15,16}
and so on...

Find the first term of the A51.

Source: GMAT Paper

RaghavSingla · Answer

A particular series is given by :

A1 = {1}
A2 = {2,3,4}
A3 = {5,6,7,8,9}
A4 = {10,11,12,13,14,15,16}
and so on...

Find the first term of the A51.

I'll give it a try.

Here each term has basically a set of integers.
Careful observation reveals 2 things:

1: Each term has last element of it its set as the square of the term number. That is for A2, say n =2 then last element of A2 is n^2=4. Similarly for A3, last element of set is 3^3=9.
2: The number of elements in the set of nth term = n+(n-1). For example for A2, n=2 and number of elements in A2 = 2+1=3. Similarly for A3, n=3 hence number of elements in set of A3 = 3+2 =5.

These 2 observations reveal that A51 will have 51+50 = 101 terms and the last term of A51 will be 51^2 and moving backwards each previous terms would be 1 less.

Hence A51 = {........,.,.,.,,.,.....,(51^2-2), (51^2-1), (51^2) }

And the first term of A51 will be 51^2-100.

Please give kudos if this helps !

RK84 · Answer

Since we need to find out the first term, we can write all the first terms as T(N)=1+2+5+10+17+.......T(51)------------------->(1)
 ..................................................... ............................... Again T(N)= 1+2+5+10+..........T(50) +T(51)---------->(2)
Subtract 2 from 1
We get 0=1+1+3+5+7+........(-)T(51), taking T(51) at the other side we have T(51)= 1+ (1+3+5+7+.....), calculate the sum to n terms of the A.P 1+ 50/2 (2+49*2)=2501 (ANSWER)..Please let me know if you have any questions.
Thanks!

KarishmaB · Answer

Note that A1 has the first positive integer. A2 has next 3, A3 has next 5, A4 has next 7 and so on...

To get the first term of A5, we discard the first 1 + 3 + 5 + 7 integers.
Similarly, to get the first term of A51, we will need to discard the first
1 + 3 + 5 + 7 + 9 + ... + (50 terms) integers

This is an AP and its sum is 
n/2(2a + (n-1)d) = (50/2) * (2*1 + 49*2) = 2500

So when we discard 2500 integers, the fist term of A51 will be 2501.

SouthCity · Answer

My attempt to make the solution simple. Note each term of the beginning of the series follows the rule An = (n-1)^2+1 n>1

A2 = (2-1)^2+1 = 2.....A4=(4-1)^2+1=10.....you can check for A3 too

A51= (51-1)^2 +1 = 50^2+1 = 2501

-SouthCity

Aamirso · Answer

A51 will have the last term as 50^2 = 2500

A51 will have the last term as 51^2 = 2601

Hence, A51 is 2501, 2502....2601

bumpbot · Answer

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Find the 51st term of the series.

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