Question

Let \(a_1, a_2, ..., a_{10}\) be an AP with common difference -3 and \(b_1, b_2, ..., b_{10}\) be a GP with common ratio 2. Let \(c_k = a_k + b_k, k = 1, 2, ..., 10\). If \(c_2 = 12\) and \(c_3 = 13\), then \(\sum_{k=1}^{10} c_k\) is equal to ___________.

Hint:

When a new sequence is defined as the sum of terms from an AP and a GP, the sum of the new sequence is simply the sum of the sums of the AP and GP. The key is to first use the given information to find the first term and common difference/ratio for each of the original sequences.

Answer 1

Correct Answer: 2021

Step 1: Set up Equations based on Given Information
For the AP: \(a_k = a_1 + (k-1)d\), where \(d=-3\). For the GP: \(b_k = b_1 r^{k-1}\), where \(r=2\). We are given \(c_k = a_k + b_k\). \(c_2 = a_2 + b_2 = (a_1 + d) + (b_1 r) = a_1 - 3 + 2b_1 = 12 \implies a_1 + 2b_1 = 15\) (1) \(c_3 = a_3 + b_3 = (a_1 + 2d) + (b_1 r^2) = a_1 - 6 + 4b_1 = 13 \implies a_1 + 4b_1 = 19\) (2)
Step 2: Solve for \(a_1\) and \(b_1\)
We have a system of two linear equations in \(a_1\) and \(b_1\). Subtract equation (1) from equation (2): \[ (a_1 + 4b_1) - (a_1 + 2b_1) = 19 - 15 \] \[ 2b_1 = 4 \implies b_1 = 2 \] Substitute \(b_1 = 2\) into equation (1): \[ a_1 + 2(2) = 15 \implies a_1 + 4 = 15 \implies a_1 = 11 \] Step 3: Calculate the Required Sum
We need to find \(\sum_{k=1}^{10} c_k\). \[ \sum_{k=1}^{10} c_k = \sum_{k=1}^{10} (a_k + b_k) = \sum_{k=1}^{10} a_k + \sum_{k=1}^{10} b_k \] This is the sum of the first 10 terms of the AP plus the sum of the first 10 terms of the GP. Sum of AP (\(S_{AP}\)): The formula is \(S_n = \frac{n}{2}[2a_1 + (n-1)d]\). Here \(n=10, a_1=11, d=-3\). \[ S_{AP} = \frac{10}{2}[2(11) + (10-1)(-3)] = 5[22 + 9(-3)] = 5[22 - 27] = 5(-5) = -25 \] Sum of GP (\(S_{GP}\)): The formula is \(S_n = \frac{b_1(r^n - 1)}{r-1}\). Here \(n=10, b_1=2, r=2\). \[ S_{GP} = \frac{2(2^{10} - 1)}{2-1} = 2(1024 - 1) = 2(1023) = 2046 \] Total Sum: \[ \sum_{k=1}^{10} c_k = S_{AP} + S_{GP} = -25 + 2046 = 2021 \] Step 4: Final Answer
The value of the sum is 2021.