Let the sum of the first n terms of a non

Question:

Let the sum of the first n terms of a non

constant A.P., $\quad a_{1}, \quad a_{2}, \quad a_{3}, \ldots \ldots$ be

$50 \mathrm{n}+\frac{\mathrm{n}(\mathrm{n}-7)}{2} \mathrm{~A}$, wherre $\mathrm{A}$ is a constant. If $\mathrm{d}$ is the common difference of this A.P., then the ordered pair $\left(d, a_{50}\right)$ is equal to

 

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

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

  3. $(50,50+46 \mathrm{~A})$

  4. $(50,50+45 \mathrm{~A})$


Correct Option: 1

Solution:

$\mathrm{S}_{\mathrm{n}}=50 \mathrm{n}+\frac{\mathrm{n}(\mathrm{n}-7)}{2} \mathrm{~A}$

$\mathrm{T}_{\mathrm{n}}=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}$

$=50 \mathrm{n}+\frac{\mathrm{n}(\mathrm{n}-7)}{2} \mathrm{~A}-50(\mathrm{n}-1)-\frac{(\mathrm{n}-1)(\mathrm{n}-8)}{2} \mathrm{~A}$

$=50+\frac{\mathrm{A}}{2}\left[\mathrm{n}^{2}-7 \mathrm{n}-\mathrm{n}^{2}+9 \mathrm{n}-8\right]$

$=50+\mathrm{A}(\mathrm{n}-4)$

$=\mathrm{A}$

$\mathrm{T}_{50}=50+46 \mathrm{~A}$

$\left(\mathrm{d}, \mathrm{A}_{50}\right)=(\mathrm{A}, 50+46 \mathrm{~A})$

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