Let the sum of the first n terms of a non-constant A.P. a1,a2,a3,…….. be 50n+n(n−7)2A where A is a constant. If d is the common difference of this A.P., then the ordered pair  d,a50 is equal to

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

1. A

2. B

3. C

4. D

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### Solution:

Given, ${S}_{n}=50n+\frac{n\left(n-7\right)A}{2}$

We know that ${a}_{n}={S}_{n}-{S}_{n-1}$

$\begin{array}{l}=50n+1\frac{n\left(n-7\right)A}{2}-50\left(n-1\right)-\frac{\left(n-1\right)\left(n-8\right)A}{2}\\ =50+\frac{A}{2}\left[{n}^{2}-7n-{n}^{2}+9n-8\right]=50+A\left(n-4\right)\end{array}$

Now, $d={a}_{n}-{a}_{n-1}=50+A\left(n-4\right)-50-A\left(n-5\right)=A$

And ${a}_{50}=50+\left(50-4\right)A=50+46A$

Hence, ordered pair $\left(d,{a}_{50}\right)=\left(A,50+46A\right)$  Register to Get Free Mock Test and Study Material

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