An algebra problem by Thomas Jacob

Algebra Level 4

The number of terms of an arithmetic progression is even.

The sum of all the odd terms and even terms of this progression are 24 and 30, respectively.

The last term exceeds the first term by 10.5

Find the total number of terms.

Clarifications: A term of the form a 2 n 1 a_{2n-1} is odd while a term of the form a 2 n a_{2n} is even where n is a natural number


The answer is 8.

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2 solutions

Thomas Jacob
Oct 28, 2016

Let us assume the number of terms in this progression to be 2 n 2n .

Then the number of odd and even terms are equal i.e. n n

The sum of the odd terms and even terms is given as 24 24 and 30 30 respectively.

We know S n S_{n} = n 2 \displaystyle \frac{n}{2} [ 2 a 2a + + ( n 1 ) (n-1) d d ]

For the odd sequence, let the first term be a a and common difference be 2 d 2d ,

d d being the common difference of the original A.P with 2 n 2n terms.

For the even sequence, the first term is equal to a + d a + d and the common difference is 2 d 2d as well.

Plugging these values into the S n S_{n} formula for odd and even terms, we get two equations.

For the odd sequence, we get,

n 2 \displaystyle \frac{n}{2} [ 2 a 2a + + ( n 1 ) (n-1) 2 d 2d ] = 24 24

This can be simplified to, a n an + + n d ( n 1 ) nd( n - 1)

For the even sequence,

n 2 \displaystyle \frac{n}{2} [ 2 ( a + d ) 2(a+d) + + ( n 1 ) (n-1) 2 d 2d ] = 30 30

This expression can be further simplified to,

( a + d ) n + n ( n 1 ) d = 30 (a + d) n + n ( n - 1) d = 30

Solving these two equations, we get, n d = 6 n d = 6

Finally, since we know the difference between the first and last terms, we can write another equation,

a n a_{n} = a + ( 2 n 1 ) d a + (2 n - 1) d

or, a + 10.5 a + 10.5 = a + ( 2 n 1 ) d a + (2 n - 1) d

or, ( 2 n 1 ) d (2n-1)d = 10.5 10.5 .

Plugging in the value of n d nd we found earlier, we get d = 1.5 d = 1.5

Now, plugging in this value of d d into n d = 6 n d = 6 , we get n = 4 n = 4

Since we wanted the number of terms in our original AP, the answer is 2 n 2n = A A = 8 8

I really like this problem, I'm certain you can contribute problems which the community would enjoy :) Here are some guidelines for improving the problem and solution:

  1. Make the problem simple for others to undestand. Clarify what "odd terms" and "even terms" mean. I believe you are referring to "odd/even-indexed terms", as opposed to "terms which are odd/even". IE if a 1 = 2 a_1 = 2 , is that an odd term or an even term?

  2. Avoid making people jump through hoops in order to answer the question. There is no reason to ask for 1000A, other than to trip them up at the last part. I've edited the question accordingly.

  3. The solution should help those who cannot solve the problem to understand what steps you took to get there. State what are the "we get two equations" and list them out. This will make it easier for others to understand how you "solved the three equations formed".

Calvin Lin Staff - 4 years, 7 months ago

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I'm glad you enjoyed the problem :) I've updated the problem and the solution. Thank you for the suggestions!

Thomas Jacob - 4 years, 7 months ago
Calvin Lin Staff
Nov 3, 2016

Suppose that there are 2 n 2n terms in the AP. Let the common difference be d d .

If we consider the odd-indexed sequence with the even-indexed sequence termwise, the difference of each pair of terms is d d , hence the difference of these two sequences is n × d = 30 24 = 6 n \times d = 30 - 24 = 6 .

We are given that the last term exceed the first term by 10.5, hence a + ( 2 n 1 ) d a = ( 2 n 1 ) d = 10.5 a+ (2n-1)d - a = (2n-1) d = 10.5 .

Solving these 2 equations by substituting d = 6 n d = \frac{6}{n} , we obtain n = 4 , d = 1.5 n = 4, d = 1.5 .

Hence, there are 2 n = 8 2n = 8 terms in the sequence.

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