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What is the value of p; if the sum of consecutive odd numbers from 1 t

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Sajjad1994
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What is the value of p; if the sum of consecutive odd numbers from 1 t [#permalink] New post   22 Oct 2020, 03:46
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57% (01:58) correct 43% (02:12) wrong based on 61 sessions
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What is the value of p; if the sum of consecutive odd numbers from 1 to p equals 225 (p being a term of the series)?

(A) 15

(B) 21

(C) 29

(D) 30

(E) 31
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C
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CrackverbalGMAT
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What is the value of p; if the sum of consecutive odd numbers from 1 t [#permalink] New post   Updated on: 22 Oct 2020, 08:23
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Sajjad1994 wrote:
What is the value of p; if the sum of consecutive odd numbers from 1 to p equals 225 (p being a term of the series)?

(A) 15

(B) 21

(C) 29

(D) 30

(E) 31



The sum of consecutive odd numbers (natural) = \(n^2\)

\(n^2 = 225 = 15^2\)

n = 15

Therefore the 15th term = 29


Option C

Arun Kumar

PS: Some important series to Note

Sum of the 1st n natural numbers = \(\frac{n(n+1)}{2}\)


Sum of squares of the 1st n natural numbers = \(\frac{n(n+1)(2n+1)}{6}\)



Sum of cubes of the 1st n natural numbers = \([\frac{n(n+1)}{2}]^2\)


Sum of the 1st n even natural numbers = n(n + 1)

Originally posted by CrackverbalGMAT on 22 Oct 2020, 07:43.
Last edited by CrackverbalGMAT on 22 Oct 2020, 08:23, edited 1 time in total.
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What is the value of p; if the sum of consecutive odd numbers from 1 t [#permalink] New post   22 Oct 2020, 08:15
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Sajjad1994 wrote:
What is the value of p; if the sum of consecutive odd numbers from 1 to p equals 225 (p being a term of the series)?

(A) 15

(B) 21

(C) 29

(D) 30

(E) 31


sum of consecutive odd numbers in series ; ( n+1)^2/ 4
here given ( p+1)^2 = 225*4
square both sides
p+1 = 30
p=29

and term of the series ; n^2 = 225 ; n= 15 i.e total odd numbers whose sum equals to 225

CrackVerbalGMAT
question says sum of odd numbers 1 to p ; where p being a term so wont the answer value be 29 ..
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Re: What is the value of p; if the sum of consecutive odd numbers from 1 t [#permalink] New post   22 Oct 2020, 08:21
The sum of first n consecutive odd integers is n^2, where n is the number of terms

so n^2=225, n=15

As mentioned above n is the number of term and not the last term. But the question asks us the value of the last term

So the the 15th term will be 29.

Answer-C
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Re: What is the value of p; if the sum of consecutive odd numbers from 1 t [#permalink] New post   22 Oct 2020, 08:29
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Edited and corrected. Thank you Archit3110 and rocky620
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Re: What is the value of p; if the sum of consecutive odd numbers from 1 t [#permalink] New post   22 Oct 2020, 10:43
The formula to find the sum of the first n consecutive odd integers is: n^2
n^2=225 n=15 So n is 15th odd number or 29. C)
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Re: What is the value of p; if the sum of consecutive odd numbers from 1 t [#permalink] New post   22 Oct 2020, 10:50
I was trying to remember the formula and I am never good at number properties formulas so used the average formula to quickly sum and get the answer.

First 7 Odds = 1 + 3+ 5+ 7+ 9 +11+13 = 7*7 = 49
Second 7 Odds = 15 + 17 + 19+ 21+ 23+ 25+ 27 = 21*7 = 147
Total so far = 196 - > Thus only remains 29 And total values are 15.

Might seem tedious but for me it was much faster than trying to think up formulas and only took 1.5Mins.
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Re: What is the value of p; if the sum of consecutive odd numbers from 1 t [#permalink]
22 Oct 2020, 10:50
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