JEE Mains · Maths · STD 11 - 8. sequence and series
The \(20^{\text {th}}\) term from the end of the progression \(20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots,-129 \frac{1}{4}\) is :-
- A \(-118\)
- B \(-110\)
- C \(-115\)
- D \(-100\)
Answer & Solution
Correct Answer
(C) \(-115\)
Step-by-step Solution
Detailed explanation
\(20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots,-129 \frac{1}{4}\) This is \(A.P\). with common difference \( d_1=-1+\frac{1}{4}=-\frac{3}{4} \) \( -129 \frac{1}{4}, \ldots \ldots \ldots . . ., 19 \frac{1}{4}, 20\) This is also \(A.P.\) \(a=-129 \frac{1}{4}\)…
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