Sequences & Series
AP — nth Term from End
nta_pyq_2024_jan
Grade 11
Question:
The $20^{\text{th}}$ term from the end of the progression $20,19\dfrac{1}{4},18\dfrac{1}{2},17\dfrac{3}{4},\ldots,-129\dfrac{1}{4}$ is:
Step-by-Step Solution
Key Concept: AP: $a=20$, $d=-3/4$. Last term $=-129\frac{1}{4}=-517/4$. 20th from end = 20th term of the reversed AP with $a'=-129 rac{1}{4}$, $d'=3/4$.
$-129\frac{1}{4}+19\times\frac{3}{4}=-129\frac{1}{4}+14\frac{1}{4}=-115$.
Correct Answer: 3