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Sequences and Series question

2024 · 27 Jan · Shift 2 · Q39

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2024 · 27 Jan · Shift 2 · Q39

Sequences and Series question

2024 · 27 Jan · Shift 2 · Q39

JEE MainMathematicsSequences and SeriesMCQ+4 / −1

\text { The } 20^{\text {th }} \text { 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} \text { is : }

A
$-115$
B
$-100$
C
$-110$
D
$-118$

View written solutionFree

Correct answer: A

Identify the progression

The given sequence is: $20,\ 19\tfrac{1}{4},\ 18\tfrac{1}{2},\ 17\tfrac{3}{4},\ \ldots,\ -129\tfrac{1}{4}$

This is an arithmetic progression (A.P.) with:

First term: $a=20$
Common difference: $d=19\tfrac{1}{4}-20=-\tfrac{3}{4}$

So, $d=-\frac{3}{4}$

Find the total number of terms

The last term is: $l=-129\tfrac{1}{4}=-\frac{517}{4}$

Using the formula for the $n$ th term of an A.P.: $a_n=a+(n-1)d$

Substitute the values: $-\frac{517}{4}=20+(n-1)\left(-\frac{3}{4}\right)$

Write $20$ as $\frac{80}{4}$ : $-\frac{517}{4}=\frac{80}{4}-\frac{3}{4}(n-1)$

Multiply by $4$ : $-517=80-3(n-1)$

$-597=-3(n-1)$

$n-1=199$

$n=200$

So the progression has 200 terms.

Find the 20th term from the end

The 20th term from the end is the: $(200-20+1)=181^{\text{st}}$ term from the beginning.

Now compute the $181^{\text{st}}$ term: $a_{181}=20+(181-1)\left(-\frac{3}{4}\right)$ $=20+180\left(-\frac{3}{4}\right)$ $=20-135$ $=-115$

Check options

A: $-115$ ✅
B: $-100$ ❌
C: $-110$ ❌
D: $-118$ ❌

Therefore, the correct answer is: $\boxed{-115}$

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