Let the sum of the first

Question:

Let the sum of the first $\mathrm{n}$ terms of a non-constant A.P., $\mathrm{a}_{1}$,

$a_{2}, a_{3}$,

be $50 n+\frac{n(n-7)}{2} A$, where $A$ is a

constant. If $\mathrm{d}$ is the common difference of this A.P., then the ordered pair $\left(\mathrm{d}, \mathrm{a}_{50}\right)$ is equal to:

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

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

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

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


Correct Option: , 4

Solution:

$\because S_{n}=\left(50-\frac{7 A}{2}\right) n+n^{2} \times \frac{A}{2} \Rightarrow \mathrm{a}_{1}=50-3 \mathrm{~S}$

$\therefore \mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=\left(\mathrm{S}_{\mathrm{n}_{2}}-\mathrm{S}_{\mathrm{n}_{1}}\right)-\mathrm{S}_{\mathrm{n}_{1}}$

$\Rightarrow \mathrm{d}=\frac{A}{2} \times 2=A$

Now, $a_{50}=a_{1}+49 \times d$

$=(50-3 \mathrm{~A})+49 \mathrm{~A}$

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

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

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