<p><strong>28.</strong> The number of terms common between the series \(1 + 2 + 4 + 8 + \cdots\) to 100 terms and \(1 + 4 + 7 + 10 + \cdots\) to 100 terms is</p>
Step-by-Step Solution
Key Concept: A term is common to both series if it can be expressed as both 2^(k-1) (geometric series) and 1+3(m-1)=3m-2 (arithmetic series). Set them equal and find which exponents k yield valid integer values of m ≤ 100.
<p><strong>Step 1:</strong> Write the general terms.<br>Geometric series: a_k = 2^(k-1) where k = 1 to 100<br>Arithmetic series: b_m = 3m - 2 where m = 1 to 100</p><p><strong>Step 2:</strong> For common terms: 2^(k-1) = 3m - 2, so m = (2^(k-1) + 2)/3</p><p><strong>Step 3:</strong> Check which powers of 2 give m as an integer.<br>• k=1: 2^0=1, m=(1+2)/3=1 ✓<br>• k=2: 2^1=2, m=(2+2)/3≈1.33 ✗<br>• k=3: 2^2=4, m=(4+2)/3=2 ✓<br>• k=4: 2^3=8, m=(8+2)/3≈3.33 ✗<br>• k=5: 2^4=16, m=(16+2)/3=6 ✓<br>• k=6: 2^5=32, m=(32+2)/3≈11.33 ✗<br>• k=7: 2^6=64, m=(64+2)/3=22 ✓<br>• k=8: 2^7=128, m=(128+2)/3≈43.33 ✗<br>• k=9: 2^8=256, m=(256+2)/3≈86 ✓<br>• k=10: 2^9=512, m=(512+2)/3≈171.33 > 100 ✗</p><p><strong>Step 4:</strong> Valid solutions where m ≤ 100: k ∈ {1,3,5,7,9}<br>Common terms: 1, 4, 16, 64, 256</p><p>∴ Answer: <strong>5</strong></p>
Correct Answer: C