Let a_1,a_2, ... ,a_n be a given A.P. whose commonn difference is an integer andn S_n = a_1 + a_2 + ... + a_n. Ifn a_1 = 1,a_n = 300 and 15 <= n <= 50, then the orderedn pair ( S_n - 4,a_n - 4 ) is equal to

(c): We know that a_n=a_1+(n-1) dn => 300=1+(n-1) d[Because a_1=1. and a_n=300 (Given) ]n => d=(299 / n-1)=(13 x 23 / n-1), which is an integer.nIt is possible if n-1= ± (13), ± (23), ± (299), ± 1nn => n=14,-12,24,-22,300,-298,2,0nnnBut n in [15,50]nnTherefore n=24 => d=(299 / 24-1)=13nnnNow, S_n-4=S_24-4=S_20=(20 / 2)[2(1)+(20-1) 13]=2490nnEE =a_24-4=a_20=a_1+19 d=1+19 x 13 n =1+247=248 Therefore (S_n-4, a_n-4)=(2490,248)n n

Let a_{1},a_{2},,a_{n} be a given A.P. whose common ... | Sequence & Series (Mathematics)

(c): We know that

a_n=a_1+(n-1) d

\Rightarrow 300=1+(n-1) d\left[\because a_1=1\right.

and

a_n=300

(Given)

]

\Rightarrow d=\frac{299}{n-1}=\frac{13 \times 23}{n-1}

, which is an integer.\nIt is possible if

n-1= \pm(13), \pm(23), \pm(299), \pm 1

\n\Rightarrow n=14,-12,24,-22,300,-298,2,0\n

\n\nBut

n \in[15,50]

\n\therefore n=24 \Rightarrow d=\frac{299}{24-1}=13\n

\n\nNow,

S_{n-4}=S_{24-4}=S_{20}=\frac{20}{2}[2(1)+(20-1) 13]=2490

\n\nEE YouTube Channel-https://www.youtube.com/@DivineJEE\n

\n\begin{aligned}\na_{n-4} & =a_{24-4}=a_{20}=a_1+19 d=1+19 \times 13 \\\n& =1+247=248 \therefore\left(S_{n-4}, a_{n-4}\right)=(2490,248)\n\end{aligned}\n

Explanation

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