Sum of first 11 terms of an arithmetic progression is 275. Which of

Question

Sum of first 11 terms of an arithmetic progression is 275. Which of the following can be the sum of the first 12 terms, if the first term and the common difference are positive integers?

A. 300
B. 310
C. 316
D. 324
E. 333

e-GMAT     Question of the Week #18

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Aman1196 · Accepted Answer

For sum of first 11 terms of the progression
we can use the formula:
n/2(2a+(n-1)d)
which will give us the value
50=2a+10d

also sum of progression is:
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sayankh · Answer

Since we are given, the sum of first 11 terms of an arithmetic progression is 275. We can take the first term to be 'a' and the subsequent terms with a common difference of 'd', added to the previous term in the sequence.

Hence, the terms would be {a, a+d, a+2d, ..., a+10d} when the series has 11 terms in it.

Similarly, when the series would be having 12 terms, the relevant terms would be {a, a+d, ..., a+11d}

Utilizing the formula of AP, we know sum of 'n' terms would be S = n/2 *

Substituting n = 11, we get the following.

S = 11/2   = 275

Simplifying this equation, we get the value of a+5d to be 25.

Now, since we are being asked to determine the sum of 12 terms in the series, we can represent it as the following.

(Sum of 11 terms in the series + 12th term)

So, this would be (275 + a + 11d). Since we know, the value of a + 5d is 25. We can represent it as, 300 + 6d.

Hence, we know the sum of 12 terms in the series would be more than 300 and also the common difference would be a multiple of 6 (it is also given that the common difference is a positive integer).

Reviewing the answer choices given, the answer would be 'D'.

EgmatQuantExpert · Answer

Solution 
  Given:  
 • The sum of first 11 terms of an arithmetic progression = 275
• First term and common difference are positive integers

To find:  
 • The sum of the first 12 terms

Approach and Working:   
 • Let us assume that the first term of the arithmetic progression is ‘a’ and common difference is ‘d’
• Sum of first 11 terms,  S_{11}  = a + (a + d) + (a + 2d) + (a + 3d) + …. + (a + 9d) + (a + 10d)
 o Implies,  S_{11}  = 11a + d * (1 + 2 + 3 + 4 + … + 9 + 10) = 11a + d * 10 * 11/2 = 11a + 55d
o We are given,  S_{11}  = 275 = 11a + 55d
o Thus, a + 5d = 25 ……… (1)

• We can write the sum of first 12 terms as  S_{12}  =  S_{11}  +  12^{th}  term
 o Implies,  S_{12}  = 275 + (a + 11d) = 275 + (a + 5d) + 6d = 300 + 6d
o We are given that d is a positive integer

• Now, if we observe the answer choices, the only option which is in the form of 300 + 6d is 324

Therefore,  S_{12}  can be equal to 324

Hence the correct answer is Option D.

Answer: D

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Kinshook · Answer

Sum of first 11 terms of an arithmetic progression is 275.

Which of the following can be the sum of the first 12 terms, if the first term and the common difference are positive integers?

Arithmetic progression: - 
First term = a
Commond difference = d
Number of terms = n
Sum of n terms = n/2{2a + (n-1)d}

Sum of 11 terms = 11/2 {2a + 10d} = 11a + 55d = 11 {a+5d} = 275
a + 5d = 25
2a + 10d = 50

12th term = a + 11d

Sum of first 12 terms = 12/2 {2a + 11d} = 6{2a + 11d} = 6(2a + 10d) + 6d = 300 + 6d

Only Option D 324 is of the form since d is a positive integer.

IMO D

vijayram24 · Answer

We get from the first statement that 11/2(2a+10d) = 275 => a+5d = 25 
We are given that a,d are integers => if a=5,d=4 , if a=10,d=3 , if a=15, d=2 (a=0,d=0 cases are omitted and all other values can be omitted as they are all negative)
For sum of 12 , we need to find 6(2a+11d) => 6(50)+6d => 300+6d.
Possible options are 318,312,324.
From the options it is 324(D)

willykuo · Answer

(a1 + a11)*11 / 2 = 275
=> a1 + a11 = 50 ---- (1)

a11 - a1 = 10d ---- (2) (since it's arithmetic progression)

(1) + (2) => a11 = 25 + 5d
a12 = a11 + d = 25 + 6d

a1 + .... a11 + a12 = 275 + 25 + 6d = 300 + 6d
and then observe options found that only 324 works

Kinshook · Answer

Sum of first 11 terms of an arithmetic progression is 275. Which of the following can be the sum of the first 12 terms, if the first term and the common difference are positive integers?

Let the first term be a and common difference be d

Sum of first 11 terms = 11/2 (2a + 10d) = 275
a + 5d = 25

Sum of first 12 terms = 12/2 (2a + 11d) = 6(2a + 11d) = 6(2a + 10d) + 6d = 12*25 + 6d = 300 + 6d

Only 324 = 300 + 6*4 is of the required form

IMO D

Ldsiam898 · Answer

Given: Sum of first 11 terms = 275
Sum formula for AP: Sₙ = n/2 ×   
S11 = 11/2 ×   = 275
⟹ 11(a + 5d) = 275
⟹ a + 5d = 25 ... (i)
Notice that a + 5d is actually the 6th term (a6 = a + 5d).
Now we need to find the sum of 12 terms:
S12 = 12/2 ×   = 6(2a + 11d)
From (i), a = 25 - 5d. Since both a and d are positive integers:
a > 0 ⟹ 25 - 5d > 0 ⟹ d < 5
So, d can be 1, 2, 3, or 4:
For d=1, a=25-5*1=20, and S12=6(2*20+11*1)=306.
d=2, a=25-5*2=15, S12=6(2*15+11*2)=312.
d=3, a=25-5*3=10, S12=6(2*10+11*3)=318
d=4, a=25-5*4=5, S12=6(2*5+11*4)=324
So S12 can be 306, 312, 318, or 324.
Given the options A. 300, B. 310, C. 316, D. 324, E. 333—only 324 matches

Sum of first 11 terms of an arithmetic progression is 275. Which of

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Sum of first 11 terms of an arithmetic progression is 275. Which of

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