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Question

Consider the following for the next items that follow:

Let a1, a2, a3 ... be in AP such that a+ a+ a10 + a15 + a20 + a25 + a30 + a34 = 300.

What is a+ a- a10 - a15 - a20 - a25 + a30 + a34 equal to ?

This question was previously asked in
NDA I 2023 GAT Previous Year Paper (16-Apr-2023)
The correct answer is

0

Understanding the Arithmetic Progression Problem

The question involves an Arithmetic Progression (AP), which is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference.

We are given a specific sum of eight terms in the AP:

\[ a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{25} + a_{30} + a_{34} = 300 \]

We need to find the value of another expression involving some of these terms:

\[ a_1 + a_5 - a_{10} - a_{15} - a_{20} - a_{25} + a_{30} + a_{34} \]

Method 1: Using the General Term Formula of AP

The general term of an AP is given by the formula:

\[ a_n = a + (n-1)d \]

where \(a\) is the first term (\(a_1\)) and \(d\) is the common difference.

Let's write out each term in the given sum using this formula:

  • \(a_1 = a + (1-1)d = a\)
  • \(a_5 = a + (5-1)d = a + 4d\)
  • \(a_{10} = a + (10-1)d = a + 9d\)
  • \(a_{15} = a + (15-1)d = a + 14d\)
  • \(a_{20} = a + (20-1)d = a + 19d\)
  • \(a_{25} = a + (25-1)d = a + 24d\)
  • \(a_{30} = a + (30-1)d = a + 29d\)
  • \(a_{34} = a + (34-1)d = a + 33d\)

Substitute these into the given sum equation:

\[ (a) + (a+4d) + (a+9d) + (a+14d) + (a+19d) + (a+24d) + (a+29d) + (a+33d) = 300 \]

Combine the terms:

\[ (a+a+a+a+a+a+a+a) + (4d+9d+14d+19d+24d+29d+33d) = 300 \] \[ 8a + (4+9+14+19+24+29+33)d = 300 \]

Summing the coefficients of \(d\): \(4+9=13, 13+14=27, 27+19=46, 46+24=70, 70+29=99, 99+33=132\).

\[ 8a + 132d = 300 \]

Dividing the entire equation by 4:

\[ 2a + 33d = 75 \]

This equation relates the first term (\(a\)) and the common difference (\(d\)).

Now consider the expression we need to evaluate:

\[ E = a_1 + a_5 - a_{10} - a_{15} - a_{20} - a_{25} + a_{30} + a_{34} \]

Substitute the general terms:

\[ E = (a) + (a+4d) - (a+9d) - (a+14d) - (a+19d) - (a+24d) + (a+29d) + (a+33d) \]

Expand and group terms with \(a\) and \(d\), paying close attention to the signs:

\[ E = a + a + 4d - a - 9d - a - 14d - a - 19d - a - 24d + a + 29d + a + 33d \]

Collect terms with \(a\):

\[ (a + a - a - a - a - a + a + a) = (1+1-1-1-1-1+1+1)a = (4-4)a = 0a = 0 \]

Collect terms with \(d\):

\[ (0d + 4d - 9d - 14d - 19d - 24d + 29d + 33d) = (4-9-14-19-24+29+33)d \] \[ = ((4+29+33) + (-9-14-19-24))d \] \[ = (66 + (-66))d = (66-66)d = 0d = 0 \]

So the expression \(E\) simplifies to:

\[ E = 0a + 0d = 0 \]

Method 2: Using a Property of Arithmetic Progressions

In an Arithmetic Progression, there is a useful property: if the sum of the indices of two terms is equal to the sum of the indices of two other terms, then the sum of those pairs of terms is also equal. That is, if \(i+j = k+l\), then \(a_i + a_j = a_k + a_l\).

Let's look at the indices of the terms in the given sum: 1, 5, 10, 15, 20, 25, 30, 34.

Consider pairs of indices:

  • \(1 + 34 = 35\)
  • \(5 + 30 = 35\)
  • \(10 + 25 = 35\)
  • \(15 + 20 = 35\)

Since the sum of indices for these pairs is the same (35), the sums of the corresponding terms are equal:

\[ a_1 + a_{34} = a_5 + a_{30} = a_{10} + a_{25} = a_{15} + a_{20} \]

Let's denote this common sum by \(K\). So, \(K = a_1 + a_{34} = a_5 + a_{30} = a_{10} + a_{25} = a_{15} + a_{20}\).

The given sum is:

\[ a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{25} + a_{30} + a_{34} = 300 \]

We can group the terms in pairs with index sums equal to 35:

\[ (a_1 + a_{34}) + (a_5 + a_{30}) + (a_{10} + a_{25}) + (a_{15} + a_{20}) = 300 \]

Substitute \(K\) for each pair sum:

\[ K + K + K + K = 300 \] \[ 4K = 300 \] \[ K = \frac{300}{4} = 75 \]

So, each pair sum is equal to 75. For example, \(a_1 + a_{34} = 75\), \(a_5 + a_{30} = 75\), etc.

Now consider the expression we need to evaluate:

\[ E = a_1 + a_5 - a_{10} - a_{15} - a_{20} - a_{25} + a_{30} + a_{34} \]

Rearrange and group the terms carefully, keeping the signs correct:

\[ E = (a_1 + a_{34}) + (a_5 + a_{30}) - (a_{10} + a_{25}) - (a_{15} + a_{20}) \]

Substitute the value \(K=75\) for each grouped pair sum:

\[ E = K + K - K - K \] \[ E = 75 + 75 - 75 - 75 \] \[ E = 150 - 150 = 0 \]

Conclusion

Both methods show that the value of the expression \(a_1 + a_5 - a_{10} - a_{15} - a_{20} - a_{25} + a_{30} + a_{34}\) is 0.

Arithmetic Progression Problem Revision

ItemDescription
Given Sum\(a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{25} + a_{30} + a_{34} = 300\)
Expression to Find\(a_1 + a_5 - a_{10} - a_{15} - a_{20} - a_{25} + a_{30} + a_{34}\)
Key AP Property UsedIf \(i+j=k+l\), then \(a_i+a_j = a_k+a_l\)
Value of Paired Sums (\(a_i+a_j\) where \(i+j=35\))\(K = 75\)
Calculated Value of Expression0

Additional Information on Arithmetic Progressions (AP)

An Arithmetic Progression (AP) is a type of sequence where the difference between any two consecutive terms is constant. This constant difference is called the common difference, usually denoted by \(d\).

  • Defining Property: \(a_{n+1} - a_n = d\) for all \(n \ge 1\).
  • First Term (\(a_1\)): The initial term of the sequence. Sometimes denoted just by \(a\).
  • Common Difference (\(d\)): The constant value added to get the next term.
  • General Term (\(a_n\)): The formula to find the \(n\)-th term is \(a_n = a_1 + (n-1)d\).
  • Sum of First \(n\) Terms (\(S_n\)): The sum of the first \(n\) terms can be found using \(S_n = \frac{n}{2}(a_1 + a_n)\) or \(S_n = \frac{n}{2}(2a_1 + (n-1)d)\).
  • Symmetric Property: In a finite AP with \(N\) terms, the sum of terms equidistant from the beginning and end is constant: \(a_k + a_{N-k+1} = a_1 + a_N\).
  • General Pair Property: As used in this problem, if \(i+j = k+l\), then \(a_i + a_j = a_k + a_l\). This holds true whether the terms are within a finite sequence or part of an infinite AP.

These properties are helpful in solving problems involving sums or relationships between non-consecutive terms in an Arithmetic Progression.

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Similar Questions

  1. The mean of 12 observations is 75. If two observation are discarded, then the mean of the remaining observations is 65. What is the mean of the discarded observations?

  2. The arithmetic mean of 1, 8, 27, 64, … up to n terms is given by

  3. Let a, b, c be in AP and k ≠ 0 be a real number. Which of the following are correct?

    1. ka, kb, kc are in AP

    2. k - a, k - b, k - c are in AP

    3. \(\frac{a}{k},\frac{b}{k},\frac{c}{k}\) are in AP

    Select the correct answer using the code given below:
  4. How many two-digit numbers are divisible by 4?

  5. If the sum of m terms of an AP is n and the sum of n terms is m, then the sum of (m + n) terms is

  6. If p 2, q 2and r 2(where p, q, r > 0) are in GP, then which of the following is / are correct?

    1. p. q and r are in GP.

    2. ln p, ln q and ln  r are in AP.

    Select the correct answer using the code given below:

  7. What is \(\displaystyle \sum_{n=1}^{34} a_n\) equal to ?

  8. What is the ratio of the first term of A to that of B ?
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Important Questions from Arithmetic Progressions

  1. The mean of 12 observations is 75. If two observation are discarded, then the mean of the remaining observations is 65. What is the mean of the discarded observations?

  2. Calculate the value of x if the arithmetic mean of the following data is zero-

    NumbersFrequency
    x + 33
    x - 77
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  3. The arithmetic and geometric means of two numbers are 65 and 25, respectively. What are these two numbers?

  4. The arithmetic mean of 1, 8, 27, 64, … up to n terms is given by

  5. The nth term of an A.P is \(\frac{{3 + {\rm{n}}}}{4}\) , then the sum of first 105 terms is

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