an=a1+(n−1)d
300=1+(n−1)d
⇒d=(n−1)299=(n−1)13×23= integer
so n−1=±13,±23,±299,±1
⇒n=14,−12,24,−22,300,−298,2,0
But n∈[15,50]⇒n=24⇒d=13
Hence,
Sn−4=S20=220[2(1)+(20−1)(13)]
⇒Sn−4=2490
And,
an−4=a20=a1+19d
=1+19×13
=248
JEE Main 2020 — Mathematics Algebra
Let a,1a2,…,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:
Held on 4 Sept 2020 · Verified 6 Jul 2026.
(2490,249)
(2480,249)
(2480,248)
(2490,248)
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