JEE Mains · Maths · STD 11 - 8. sequence and series
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 \(\left( S _{ n -4}, a _{ n -4}\right)\) is equal to
- A \((2480,249)\)
- B \((2490,249)\)
- C \((2490,248)\)
- D \((2480,248)\)
Answer & Solution
Correct Answer
(C) \((2490,248)\)
Step-by-step Solution
Detailed explanation
\(\quad a_{n}=a_{1}+(n-1) d\) \(\Rightarrow 300=1+(n-1) d\) \(\Rightarrow \quad(n-1) d=299=13 \times 23\) since, \(n \in[15,50]\) \(\therefore n=24\) and \(d=13\) \(a_{n-4}=a_{20}=1+19 \times 13=248\) \(\Rightarrow a_{n-4}=248\) \(S_{n-4}=\frac{20}{2}\{1+248\}=2490\)
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