Question

5, 11, 21, 43, 85, ?

• A. 185
• B. 170
• C. 171
• D. 181

Hint:

Look for alternating \(\pm 1\) after a common multiplier.

Answer 1

The Correct Option is C

Step 1: Sums of consecutive terms: \(5{+}11=16\), \(11{+}21=32\), \(21{+}43=64\), \(43{+}85=128\) — each is double the last, i.e. powers of \(2\).

Step 2: The next sum should be \(256\), so the missing term is \(256-85\).

Step 3: That gives \(171\).
\[ \boxed{171} \]