Let the sum of the first n terms of a non-constant A.P., a_1,a_2,a_3, ... ... .. 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,a_50 ) is equal to

a) : Given, S_n=50 n+(n(n-7) A / 2) We know that a_n=S_n-S_n-1 \ =50 n+(n(n-7) A / 2)-50(n-1)-((n-1)(n-8) A / 2) =50+(A / 2)[n^2-7 n-n^2+9 n-8]=50+A(n-4) \ Now, d=a_n-a_n-1=50+A(n-4)-50-A(n-5)=A And a_50=50+(50-4) A=50+46 A Hence, ordered pair (d, a_50)=(A, 50+46 A)

Let the sum of the first n terms of a non-constant A.P.... | Sequence & Series (Mathematics)

a) : Given, $S_n=50 n+\frac{n(n-7) A}{2}\\$ We know that $a_n=S_n-S_{n-1}$ $\\ \begin{aligned} & =50 n+\frac{n(n-7) A}{2}-50(n-1)-\frac{(n-1)(n-8) A}{2} \\& =50+\frac{A}{2}\left[n^2-7 n-n^2+9 n-8\right]=50+A(n-4) \end{aligned}\\$ Now, $d=a_n-a_{n-1}=50+A(n-4)-50-A(n-5)=A\\$ And $a_{50}=50+(50-4) A=50+46 A\\$ Hence, ordered pair $\left(d, a_{50}\right)=(A, 50+46 A)$

Explanation

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