Let a1, a2 ldots, an be a given A.P. whose common difference is an integer and S n = a 1+ a 2+ ldots+ a n If a1=1, an=300 and 15 ≤ n ≤ 50, then the ordered pair ( S n -4, a n -4) is equal to

Question

Let a1, a2 ldots, an be a given A.P. whose common difference is an integer and S n = a 1+ a 2+ ldots+ a n If a1=1, an=300 and 15 ≤ n ≤ 50, then the ordered pair ( S n -4, a n -4) is equal to (A) (2480,249) (B) (2490,249) (C) (2490,248) (D) (2480,248)

Tardigrade Admin · Accepted Answer

an =a1+(n-1) d ⇒ 300=1+(n-1) d ⇒ (n-1) d=299=13 × 23 text since, n ∈[15,50] ∴ n =24 text and d=13 an-4=a20=1+19 × 13=248 ⇒ an-4=248 S n-4=(20/2) 1+248 =2490

Q. 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

(2480,249)

(2490,249)

(2490,248)

(2480,248)

$a_{n} = a_{1} + (n - 1) d$
$\Rightarrow 300 = 1 + (n - 1) d$
$\Rightarrow (n - 1) d = 299 = 13 \times 23$
$since, n \in [15, 50]$
$∴ n = 24 and d = 13$
$a_{n - 4} = a_{20} = 1 + 19 \times 13 = 248$
$\Rightarrow a_{n - 4} = 248$
$S_{n - 4} = \frac{20}{2} {1 + 248} = 2490$

Q. Let a1​,a2​…,an​ be a given A.P. whose common difference is an integer and Sn​=a1​+a2​+…+an​ If a1​=1,an​=300 and 15≤n≤50, then the ordered pair (Sn−4​,an−4​) is equal to