Let $a_{1}, a_{2}, \ldots a_{n}$ be a given A.P. whose common difference is an integer and $S_{n}=a_{1}+a_{2} +\ldots+a_{n}$. If $a_{1}=1, a_{n}=300$ and $15 \leq n \leq 50$, then the ordered pair ( $S_{n-4}, a_{n-4}$ ) is equal to:

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Sequences and Series

Let

a_{1}, a_{2}, \dots a_{n}

be a given A.P. whose common difference is an integer and

S_{n} = a_{1} + a_{2} + \dots + a_{n}

. If

a_{1} = 1, a_{n} = 300

and

15 \leq n \leq 50

, then the ordered pair (

S_{n - 4}, a_{n - 4}

) is equal to:

Answer: (A)

Reading the given data as

a_{1} = 1

and

a_{n} = 300

, the common difference satisfies

(n - 1) d = 299

. Since

d

is an integer and

15 \leq n \leq 50

n - 1

must be a divisor of

299 = 13 \times 23

lying in that range, which forces

n - 1 = 23

, so

n = 24

and

d = 13

. Then

a_{n - 4} = a_{20} = 1 + 19 \times 13 = 248

and

S_{20} = \frac{20 ( 1 + 248 )}{2} = 2490

, giving the ordered pair

(2490, 248)

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option