In a certain sequence, each term is 7 greater than the term before it.

Question

In a certain sequence, each term is 7 greater than the term before it. If the third term of the sequence is 39, what is the 13th term?

A 102
B 109
C 116
D 123
E 130

Archit3110 · Answer

a1=x
a2=x+7
a3=x+14
.
.
.
a13=x+84

given a3= 39
39=x+14
x=25
a13= 25+84 = 109

or use formula
an=a+(n-1)*d
a13=25+ ( 13-1) * 7
a13= 25 + 84
a13= 109
IMO B

KanishkM · Answer

Above  part signifies that the series will be an AP with d = 7 and in AP =>  a_n = a + (n-1) d

a => First term , d => common difference

a_3  will be => a_3 = a + 2d 
=>  39 - 14 = a_3 
=>  a_3 = 25

a_{13}  = 25 + 12*7 = 109

Answer B

jfranciscocuencag · Answer

13 - 3 = 10

10*7 =70

39 + 70 = 109

manggarai · Answer

is this way correct:

1. calculate number of terms between S13 and S3. 13-3+1=11
2. calculate S33 using S33=a+(n-1)d, where a=39, d=7, n=11; S33=39+7(11-1)=39+70=109

?

KanishkM · Answer

Hey  manggarai

Hope you are doing great.

Point 2 has some discrepancy, said that, value of a is incorrect, it should be 25 rather than 39

In the question we have given the value of 3rd term and not the first, so to calculate the 3rd term since the set was not that difficult, you can easily calculate "a" as 
 a_3  = a + 2*7 
a = 25

But before i go forward isn't the question asking what is  a_{13}  ?

Would like to hear your thoughts on this

manggarai · Answer

Does A needs to be the first term? Can it not be the first term where N is concerned? I used s3=39 as A; Sn=S13; N= Number of Terms between S13 and S3 inclusive (13-3+1=11); so S13=39+(11-1)(7)= 39+10*7=39+70=109.

I also don't understand why S3 should be 25 if the question is already stating that s3=39.

KanishkM · Answer

Hi manggarai Inline are my answers to your questions Yes, A needs to be the first term and No it cannot be the first term where N is concerned? Some theory can be understood here, one of our gmatclub member did a good job in explaining the concept https://gmatclub.com/forum/arithmetic-p ... 61976.html Actually this is a standard formula period a_n = a + (n - 1) d a = first term d = common difference n = number of term in the series Now a_3 is given as 39, d = 7, we need to find the first term, which is "a" you can find the first term as a = 39 - 14 = 25 Now a_{13} = a + 12 * 7 = 25 + 84 = 109 Some concepts you have mixed here, knowing the number of terms wouldn't have been beneficial in this example, that was not required. Let me know if you still have a doubt

manggarai · Answer

Hi KanishkM , thanks for your response

What is the difference if I used the standard formula for the two scenarios? 1. when a is used as third term (s3), s13 is target, n=11; s13=39+7(11-1)=109 2. when a is used as first term (s1), s13 is target, n=13; s13=25+7(13-1)=109 i also have a problem with 2. because it takes time to calculate 25. whereas for 1, you can use what is given from the question.

KanishkM · Answer

The only difference between both the approaches, is that number 2 is the correct one, and i would kindly request you to unlearn the approach used in 1. (I hope i am not coming out as rude, but i still would insist you to not follow that approach.) Now talking about point 2, in which you are facing a time crunch, To improve their you can try to work on 500 level questions, While solving those try to take as much time you can take to solve them. Once you gain confidence, I doubt you will face any challenge. Now whenever you feel an approach is difficult to understand, try with smaller numbers Example been -> What is the 4th term ? When the 2nd term is 4 and common difference is 2 ? Now how will you solve this ?(this is not a GMAT question, this is just an example for learning purpose) Now since the d is small, i can easily calculate without using the formula, 4th term will be 8 How can we get 8, now if i solve it using the formula, I can easily see that the first term would had been 2 as the common difference is 2, therefore the series was 2 4 6 8 Now if i had to calculate 15th term, using the formula a_n = a + n-1 * d a = 2 d = 2 a15 is nothing but, a + 14d = 2 + 14*2 = 30(15*2) You have to learn how you can use the given data, to find the variables of the expression Case been if i was given the 10th term as 20, and the common difference d was given as 2, how would i have calculated the 50th term ?? Since i know that 10th term is nothing but a + 9d = 20 a = 20 -18 a = 2 (First term) I got the value of a, use it in the main formula or you can even write A50 = a + 49d Giving you the value as 2 + 49*2 = 100 We need to put some efforts to score high on the exam

Now i think you should get a clarity on the above logic, if you still face a challenge, i would request you to go through the theory of AP and understand what each variable stands for.

manggarai · Answer

Thanks KanishkM for your explanation. I used my approach for your examples (a1=2, d=2) but since we got the same answers (a10=20; a50=100) I have not found a reason to not use it. My guess is that my approach can be used as long as "N" corresponds to the correct number of integers between "an" and "a" in an=a+(n-1)(d).

Could we get the opinion from Bunuel chetan2u VeritasKarishma ScottTargetTestPrep?

does a in an=a+(n-1)(d) needs to be the first term a1 only or can it be something else say a3,a4 and so on?

Could we get the opinion from  Bunuel   chetan2u   VeritasKarishma   ScottTargetTestPrep ?

does a in an=a+(n-1)(d) needs to be the first term a1 only or can it be something else say a3,a4 and so on?

KanishkM · Answer

Hey manggarai No worries

Can you please try that in this example a4 = 97, d = 12, what will be a98 here ? Will you put value of a as 97 ? Did you try that approach in other questions as well ? I hope the experts will be able to explain in a precise way. Good luck

chetan2u · Answer

Hi,

when we are using thsi formula, we are generally talking of terms a to  a_n , and (n-1) is the number of gap, where each gap is d..

Now, can you use it the way you have used it, why not? Only thing you should be careful is the number of gaps (n-1). 
Say an AP is 3, 5, 7, 9, 11, 13 so  a_n=a+(n-1)d....a_6=3+(6-1)*2=3+10=13 , but say you have taken a as the 3rd term, then  a_6=a_3+(n-1)d....a_6=7+(4-1)*2=7+6=13 .. n is 4 because the terms are  a_3, a_4, a_5, a_6 , so 4 terms.

manggarai · Answer

thank you  chetan2u

KanishkM  i got a98=1225, i used a4=97 and i didnt bother to calculate a1 cause i dont think i need to? what did you get for a98?

KanishkM · Answer

Now manggarai I believe your approach is apparently reducing the difference between two terms, which is hindsight is enabling you to get to the right answer, but if you are able to use the same approach in the questions of higher difficulty then you can follow this approach, said that have a look here If you use this formula a_n = a + (n-1) d Your approach will give me the answer as, a98 = 97 + 94 * 12 = 1225, now since you will be removing some values from the series by using your logic of calculating the number of terms as 94 ( 97-4+1), you actually removed 4 terms in this process automatically giving you the correct answer each time. I have would say this is a Good logic , see if this helps you to solve higher difficulty questions or not, if it does help you, only then stick to this approach. If it doesn't, then come back to the original approach, in which you need not calculate the number of terms every time( you would be wasting precious seconds doing that) Just calculate the first term reduce the value of n by 1 and we are good with this question of lower difficulty . Thank you for having a good discussion with me

ScottTargetTestPrep · Answer

It is actually possible to make everyone happy by using the very formula a_n = a + (n - 1)d without having to find the first term.

Using the same example of s_3 = 39 and d = 7, define a new sequence b_n = s_(n + 2); in other words, b_1 = s_3 = 39, b_2 = s_4 = 46 etc. Now, b_n is also an arithmetic sequence with common difference 7, so we can apply the above formula with first term = 39, common difference = 7. We will be looking for the term s_13 = b_11. We will get b_11 = 39 + 10*7 = 109.

Above approach will work for any example you wish to test. It relies on the fact that if you omit first few elements of an arithmetic sequence, what remains is also an arithmetic sequence with the same common difference but with a different first term.

In a certain sequence, each term is 7 greater than the term before it.

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In a certain sequence, each term is 7 greater than the term before it.

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