Let the sum of the first n terms of a non-constant A.P., a 1 , a2 , a3 , ………….. be 50n + n(n 7) 2 – A, where A is a constant. If d is the common difference of this A.P., then the ordered pair (d, a50) is equal to:

Question:

Let the sum of the first $n$ terms of a non-constant A.P., $a_{1}, a_{2}, a_{3}, \ldots \ldots \ldots \ldots .$ be $50 n+\frac{n(n-7)}{2} A$, where $A$ is a constant. If $d$ is the common difference of this A.P., then the ordered pair $\left(\mathrm{d}, \mathrm{a}_{50}\right)$ is equal to: $\quad$ 

  1. $(50,50+46 \mathrm{~A})$

  2. $(50,50+45 \mathrm{~A})$

  3. $(\mathrm{A}, 50+45 \mathrm{~A})$

  4. $(\mathrm{A}, 50+46 \mathrm{~A})$


Correct Option: 4

JEE Main Previous Year 1 Question of JEE Main from Mathematics Sequences and Series chapter.
JEE Main Previous Year April 09, 2019 (I)

Solution:

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