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Let the sum of the first $n$ terms of a non-constant A.P., $a_{1},a_{2},a_{3},.....$ be $50n+\frac{n\left ( n-7 \right )}{2}A,$ whrere $A$ is a constant.

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

Option 1)
$\left ( 50,50+46A \right )$

Option 2)
$\left ( 50,50+45A \right )$

Option 3)
$\left ( A,50+45A \right )$

Option 4)
$\left ( A,50+46A \right )$

Answers (1)

best_answer

Given , $S_{n}=50n+\frac{n\left ( n-1 \right )}{2}A$ ...................(1)

$S_{n-1}=50_{\left ( n-1 \right )}+\frac{\left ( n-1 \right )\left ( n-2 \right )}{2}A$ .........................(2)

$T_{n}=S_{n}-S_{n-1}=50+A\left ( n-4 \right )$

$T_{1}=50+\left ( -3A \right )$

$T_{2}=50+\left ( -2A \right )$

$d=A$

$T_{50}=50-3A+\left ( 50-1 \right )A$

$=50+46A$

Option 1)

$\left ( 50,50+46A \right )$

Option 2)

$\left ( 50,50+45A \right )$

Option 3)

$\left ( A,50+45A \right )$

Option 4)

$\left ( A,50+46A \right )$

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