Arithmetic Progressions
NCERT Exemplar Ch 05
CBSE_NCERT_EXEMPLAR_CH05
Grade 10
Question:
If the first, second and last terms of an AP are $a, b$ and $c$ respectively, then its sum is:
$\dfrac{(a + c)(b + c - 2a)}{2(b - a)}$
$\dfrac{(b + c - 2a)}{2(b - a)}$
$\dfrac{(a + c)(b + c - 2a)}{b - a}$
$\dfrac{(a + c)(b - a)}{2(b + c - 2a)}$
Step-by-Step Solution
Key Concept: Common difference $d = b - a$. Total terms $n = 1 + \dfrac{c - a}{b - a} = \dfrac{b + c - 2a}{b - a}$. $S_n = \dfrac{n}{2}(a + c)$.
Stepwise Solution:
$d = b - a$. Last term $c = a + (n-1)(b-a) \Rightarrow n = \dfrac{c - a}{b - a} + 1 = \dfrac{b + c - 2a}{b - a}$. [0.5 Mark]
Sum $S_n = \dfrac{n}{2}(a + c) = \dfrac{(a + c)(b + c - 2a)}{2(b - a)}$. [0.5 Mark]
Marking Scheme:
• Finding total terms $n$: 0.5 Mark
• Applying sum formula: 0.5 Mark
Correct Answer: $\dfrac{(a + c)(b + c - 2a)}{2(b - a)}$
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