The problem focuses upon concepts rather than formula i.e. There is no need to use any formula for this question. In order to solve this , we must know one property of AP which is
1) Any term is equal to half the sum of the terms which are equidistant from it.
or
2) The sum of terms equidistant from the beginning & end is always same and is equal to the sum of the first & last term.

Now back to the problem:- As the series contain 23 terms, the mean of the series will the middle term of the series which is 12th term.
12th term is equal distance from both 10th & 12th term
Thus 12th term = average of 10th & 14th term = 94/2 = 47

a10 = a1 + 9d

a14 = a1 + 13d

=> a10 + a14 = 2 * a1 + 22d = 94



Now, a1 + a23 = a1 + [a1 + (23-1)d]

=> a1 + a23 = a1 + a1 + 22d

=> a1 + a23 = 2 * a1 + 22d = 94

So AM from a1 to a23 = 94/2 = 47 (Because it's an AP)

Answer - A

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10th term:
t10 = a1 + (10-1)d
t10 = a1 + 9d

14th term:
t14 = a1 +13d

t10+t14=94
2a1+22d=94


Sum of 23 terms of sequence= 23/2 [2a1+(23-1)d]
=23/2 [2a1+22d]
=23/2 * 94
=1081

Arithmetic mean = sum of 23 terms / 23 = 1081/23 =47
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mean = (first term + last term)/2

first term = a

last term = 23rd = a + (n-1)d where d is the common difference.

thus 23rd term = a + 22d

now sum given is
a+9d + a+ 13d = 2a + 22d = 94

thus mean = [a + (a+(n-1)d) ]/2 = (2a+ 22d)/2 = 94/2 = 47

Arithmetic mean from 1st to 23rd term would be the number in 12th term:

(23+1)/2=12

We just need to find the 12th term.

10,11,12,13,14

We know that the 12th term is midway from 10th and 14th. And 12th term would be (10th+14th)/2=94/2=47

Ans: "A"

OK this is how you solve this question in under a minute.

We are asked what is (s1+s23)/2

s23 = s1+22k, where k is a constant

So we need to find (2s1+22k)/2 or s1+11k

We are given s10+s14=94
So s1+9k+s1+13k=94
2s1+22k=94
s1+11k=47

So 47 is our answer
Hence, (A)

Hope it helps
Cheers!
J

QUESTION : HOW DO WE KNOW WE NEED TO FIND (2S1 + 22K ) / 2 , what manipulation was done to figure this out?

We can directly predicate it as question mention difference is constant so mean = median
let see
adding the value of first and last term/2 ,second first and second last/2....... - we can get mean.
so
Values of any - 1+23/2,
2+22/2,
3+21/2,
4+20/2,
5+19/2,
6+18/2,
7+17/2,
8+16/2,
9+15/2,
10+14 = 94/2(given)=47(mean)
mean =median =47 so answer is 47.

Let the numbers be as follows:

x, x+y, x+2y, x+3y....................................................... x+22y

1st, 2nd ..............................................................................23rd

10th number = x+9y

14th number = x+13y

Given that x+9y+x+13y = 94

2x + 22y = 94

x + 11y = 47

Note that x+11y is the middle term of the series which will also be the mean of the series.

Answer = A

Most of the previous analyses seem way too complicated.

Here's mine:
- we're given that members of the set are evenly spaced, so it's an evenly-spaced set.
- in an evenly-spaced (ordered) set the mean of any two members that are equidistant from the center is equal to the mean value of all members of the set.
- the 10th and 14th members are equidistant from the center, which is the 12th member.
- the mean of member 10 and member 14 = 47 = mean of the set. --> A.

Very simple one to explain.
Sum of 10 th term and 14 th term is a+9d + a+13d= 2a+ 22 d = 94.
Therefore a + 11d = t(12 term)=47.
Mean of arithmetic progression is sum of first and last term /2.
Therefore sum of 1st and 23 rd /2= 12 term = 47.

Superb Question
Here we need to use the basic principle involving mean and AP series
Mean of any AP series = Median
Here N=23
Median = 12th term = A+11D fro first term being A and D being the common Difference
Now Given A10 + A14 =94 => A+9D+A+13D = 2A+22D=94 => A+11D=47
Hence Median = 47
So the mean = 47

Smash that A
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Average of 23 terms should be the 13th term in the series.
Now to find the 13th term we can either add t1+t23 or as given in problem statement t10+t14 are also equi distant to LHS & RHS respectively of the t13.
Hence avg of t10+t14=47 which should also be the 13th term and hence the avg of Sum of 1st 23 terms in the series.
Option A.

the sum of any two terms in an arithmetic progression will equal twice the term exactly between them
thus the 12th term=94/2=47=mean in a 23 term progression
A

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