Problem 30 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The sum of the first 20 terms common between the series 3 \(+7+11+15+\ldots \ldots . . .\) and \(1+6+11+16+\ldots \ldots\), is |Online April 11, 2014] (a) 4000 (b) 4020 (c) 4200 (d) 4220

Step-by-Step Solution

Answer

The sum is 4020, option (b).

Step 1: Identify Common Terms

The given series are arithmetic progressions (APs) with the first series having the first term \(a_1 = 3\) and common difference \(d_1 = 4\), and the second series having the first term \(a_2 = 1\) and common difference \(d_2 = 5\). To find the intersection, we set up the equations of terms in the series to be equal:\[ 3 + 4n_1 = 1 + 5n_2 \]

Step 2: Find General Term Equations

From Step 1, we have:\[ 3 + 4n_1 = 1 + 5n_2 \] \[ 4n_1 - 5n_2 = -2 \]. This equation helps us find the common terms.

Step 3: Solve for Common Terms

To simplify, rearrange for \(n_1\):\[ 4n_1 = 5n_2 - 2 \]\[ n_1 = \frac{5n_2 - 2}{4} \]Solving for integer values, check for values of \(n_2\) that satisfy this equation yielding integer \(n_1\). Start with small \(n_2\) like 1, 2, etc.

Step 4: Find First Common Term (Both series)

By trial, when \(n_2 = 2\), we find that \(n_1 = 2\) is an integer. Find the value of that term:\[ a = 1 + 5 \times 2 = 11 \]Therefore, \(11\) is a common term.

Step 5: General Formula for Common Terms

The common term found is \(11\). We establish a general formula for common terms starting from \(11\). Let common difference be \(d = 20\) because terms align every 20 intervals:New series for common terms: \( T_k = 11 + (k-1) \times 20 \).

Step 6: Use General Formula to Calculate First 20 Common Terms

Calculate the first term and the 20th term of this new series:\[ T_1 = 11 \]\[ T_{20} = 11 + 19 \times 20 = 391 \]. Calculate the sum.

Step 7: Calculate Sum of First 20 Common Terms

The sum of the first 20 terms of the arithmetic sequence is given by the formula:\[ S_n = \frac{n}{2} \times (a + l) \]where \(a = 11\), \(l = 391\), and \(n = 20\):\[ S_{20} = \frac{20}{2} \times (11 + 391) = 10 \times 402 = 4020 \]

Step 8: Conclusion

The sum of the first 20 common terms is 4020. Therefore, the answer is (b) 4020.

Key Concepts

Common Terms

When two arithmetic progressions are given, you can find the common terms by equating their general term equations. For example, if the first sequence has a first term \( a_1 \) and a common difference \( d_1 \), and the second sequence has a first term \( a_2 \) and a common difference \( d_2 \), then the equations for their terms would be:

• First Sequence: \( a_1 + n_1 imes d_1 \)
• Second Sequence: \( a_2 + n_2 imes d_2 \)

You may find common terms by setting the two sequences equal to each other, forming an equation like: \[ a_1 + n_1 imes d_1 = a_2 + n_2 imes d_2 \]You then solve for one of the variables. In our exercise, this resulted in the equation \( 4n_1 - 5n_2 = -2 \), which, upon solving, yielded integer solutions for specific values of \( n_2 \).

Intersection of Series

The intersection of two sequences refers to the terms that appear in both sequences. In mathematics, such terms are often called common terms, and they can be found by solving the general term equation of both sequences simultaneously. For example, in our stated exercise:

• The first sequence follows: \( T_1 = 3 + 4n_1 \)
• The second sequence is: \( T_2 = 1 + 5n_2 \)

By equating the terms, as described above, the two sequences can share values. Solving these equations eventually leads to one integer solution for both sequences, marking these as common terms. Once the starting common term is identified, the subsequent common terms can be determined by following the discovered common difference, which in this situation observed was 20.

Sum of Arithmetic Sequence

The sum of an arithmetic sequence can be calculated using a specific formula. This is particularly useful once you have identified or generated a continuous series of numbers. For a sequence with \( n \) terms, an initial term \( a \), and a final term \( l \), the sum \( S_n \) becomes: \[ S_n = \frac{n}{2} \times (a + l) \] In our exercise, to find the sum of the first 20 common terms, we had:

• \( a = 11 \), the first common term
• \( l = 391 \), the 20th term
• \( n = 20 \), the total number of terms

Substituting these into the formula gave us \( S_{20} = 4020 \). This sum represents the total of the initial n terms identified as common to both sequences, ensuring that calculations hold true by confirming the placement and values of each term. Every formula application within arithmetic sequences assists in transforming complex sequences into comprehensible values.