Let \( S_1 = \{100, 105, 110, 115, \dots\} \) and \( S_2 = \{100, 95, 90, 85, \dots \} \) be two series in arithmetic progression. If \(a_k\) and \(b_k\) are the k-th terms of \(S_1\) and \(S_2\), respectively, then \( \sum_{k=1}^{20} a_k b_k \) equals ________
Show Hint
- 138250
- 137275
- 135375
- 137225
The Correct Option is A
Solution and Explanation
The problem asks for the sum of the product of corresponding terms of two arithmetic progressions. We need to find the general formula for the k-th terms, find the product, and then use summation formulas for powers of k.
Step 2: Key Formula or Approach:
1. The k-th term of an AP is \( T_k = a + (k-1)d \), where 'a' is the first term and 'd' is the common difference. 2. Find the expressions for \(a_k\) and \(b_k\). 3. Find the expression for the product \( a_k b_k \). 4. Use the formulas for the sum of series: \( \sum_{k=1}^{n} c = cn \) \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \) \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \)
Step 3: Detailed Explanation:
For series \(S_1\): First term \( a_{S1} = 100 \), common difference \( d_1 = 5 \). The k-th term is \( a_k = 100 + (k-1)5 = 100 + 5k - 5 = 95 + 5k \). For series \(S_2\): First term \( b_{S2} = 100 \), common difference \( d_2 = -5 \). The k-th term is \( b_k = 100 + (k-1)(-5) = 100 - 5k + 5 = 105 - 5k \). Now, find the product \( a_k b_k \): \[ a_k b_k = (95 + 5k)(105 - 5k) \] We can factor out 5 from each term: \[ a_k b_k = 5(19 + k) \cdot 5(21 - k) = 25(19 + k)(21 - k) \] Expand the product in the parenthesis: \[ (19 + k)(21 - k) = 19 \times 21 - 19k + 21k - k^2 = 399 + 2k - k^2 \] So, \( a_k b_k = 25(399 + 2k - k^2) \). Now we need to compute the sum \( \sum_{k=1}^{20} a_k b_k \): \[ \sum_{k=1}^{20} 25(399 + 2k - k^2) = 25 \left[ \sum_{k=1}^{20} 399 + 2\sum_{k=1}^{20} k - \sum_{k=1}^{20} k^2 \right] \] Let's calculate each sum with n=20:
\( \sum_{k=1}^{20} 399 = 399 \times 20 = 7980 \)
\( \sum_{k=1}^{20} k = \frac{20(20+1)}{2} = \frac{20 \times 21}{2} = 210 \)
\( \sum_{k=1}^{20} k^2 = \frac{20(20+1)(2 \times 20+1)}{6} = \frac{20 \times 21 \times 41}{6} = 10 \times 7 \times 41 = 2870 \)
Substitute these values back into the main expression: \[ \text{Sum} = 25 [7980 + 2(210) - 2870] \] \[ \text{Sum} = 25 [7980 + 420 - 2870] \] \[ \text{Sum} = 25 [8400 - 2870] \] \[ \text{Sum} = 25 [5530] \] \[ \text{Sum} = 138250 \] Step 4: Final Answer:
The value of \( \sum_{k=1}^{20} a_k b_k \) is 138250.
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