Arithmetic Progressions
NCERT Exemplar
CBSE
Grade 10
Question:
The $4^{\text{th}}$ term from the end of the AP: $-11, -8, -5, \dots, 49$ is:
(a) $37$
(b) $40$
(c) $43$
(d) $58$
Step-by-Step Solution
Key Concept: $n^{\text{th}}$ term from the end $= l - (n - 1)d$, where $l$ is the last term and $d$ is common difference.
Here $l = 49$, $d = -8 - (-11) = 3$. [0.5 Mark]
$4^{\text{th}}$ term from end $= l - 3d = 49 - 3(3) = 49 - 9 = 40$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Identifying last term $l=49$ and $d=3$: 0.5 Mark
Calculating $l - 3d = 40$: 0.5 Mark
Correct Answer: $40$
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