In an arithmetic progression the difference between the any : Problem Solving (PS)

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19 Jul 2011, 05:51

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kannn

In an arithmetic progression the difference between the any two consecutive terms is a constant. What is the arithmetic mean of all of the terms from the first to the 23rd in an arithmetic progression if the sum of the 10th and 14th terms of the sequence is 94?

A. 47
B. 63
C. 55
D. 94
E. It can't be determined

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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19 Jul 2011, 05:52

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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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19 Jul 2011, 09:27

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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

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11 Dec 2013, 11:17

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kannn

In an arithmetic progression the difference between the any two consecutive terms is a constant. What is the arithmetic mean of all of the terms from the first to the 23rd in an arithmetic progression if the sum of the 10th and 14th terms of the sequence is 94?

A. 47
B. 63
C. 55
D. 94
E. It can't be determined

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

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04 Jul 2014, 09:30

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jlgdr

kannn

In an arithmetic progression the difference between the any two consecutive terms is a constant. What is the arithmetic mean of all of the terms from the first to the 23rd in an arithmetic progression if the sum of the 10th and 14th terms of the sequence is 94?

A. 47
B. 63
C. 55
D. 94
E. It can't be determined

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?

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07 Oct 2014, 18:35

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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.

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08 Oct 2014, 00:28

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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

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09 Mar 2016, 10:10

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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.

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09 Mar 2016, 11:09

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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.

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06 Aug 2016, 00:42

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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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07 Aug 2017, 00:53

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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.

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kannn

In an arithmetic progression the difference between the any two consecutive terms is a constant. What is the arithmetic mean of all of the terms from the first to the 23rd in an arithmetic progression if the sum of the 10th and 14th terms of the sequence is 94?

A. 47
B. 63
C. 55
D. 94
E. It can't be determined

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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13 Mar 2022, 08:43

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Given: In an arithmetic progression the difference between the any two consecutive terms is a constant.

Asked: What is the arithmetic mean of all of the terms from the first to the 23rd in an arithmetic progression if the sum of the 10th and 14th terms of the sequence is 94?

For an arithmetic progression:-
$t_n = a + (n-1)d$
a = first term
d = common difference

The arithmetic mean of all of the terms from the first to the 23rd in an arithmetic progression = 23/2 (2a + 22d)/23 = a + 11d

The sum of the 10th and 14th terms of the sequence is 94.
a + 9d + a + 13d = 94
2a + 22d = 94
a + 11d = 47

IMO A

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