Find the sum of all two-digit odd positive integers.
Step-by-Step Solution
Key Concept: AP is $11, 13, 15, \dots, 99$. First term $a = 11$, $d = 2$, last term $l = 99$.
First term $a = 11$, common difference $d = 2$, last term $a_n = 99$.
$99 = 11 + (n - 1)(2) \Rightarrow 2(n - 1) = 88 \Rightarrow n - 1 = 44 \Rightarrow n = 45$. [1.0 Mark]
Sum $S_{45} = \dfrac{45}{2}(11 + 99) = \dfrac{45}{2}(110) = 45 \times 55 = 2475$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding total terms $n = 45$: 1.0 Mark
Calculating sum $S_{45} = 2475$: 1.0 Mark
Correct Answer: