Let the sum of the first n terms of a non-con | vedclass

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Question

${S_n} = 50n + \frac{{n\left( {n - 7} \right)}}{2}A$

${T_n} = {S_n} - {S_{n - 1}}$

$ = 50n + \frac{{n\left( {n - 7} \right)}}{2}A - 50\left( {n

Vedclass · Accepted Answer

$(A, 50 + 46A)$

Vedclass · Answer

${S_n} = 50n + \frac{{n\left( {n - 7} \right)}}{2}A$

${T_n} = {S_n} - {S_{n - 1}}$

$ = 50n + \frac{{n\left( {n - 7} \right)}}{2}A - 50\left( {n - 1} \right) - \frac{{\left( {n - 1} \right)\left( {n - 8} \right)}}{2}A$

$ = 50 + \frac{A}{2}\left[ {{n^2} - 7n - {n^2} + 9n - 8} \right]$

$ = 50 + A\left( {n - 4} \right)$

$d = {T_n} - {T_{n - 1}}$

$ = 50 + A\left( {n - 4} \right) - 50 - A\left( {n - 5} \right)$

$ = A$

${T_{50}} = 50 + 46A$

$\left( {d,{A_{50}}} \right) = \left( {A,50 + 46A} \right)$

Let the sum of the first $n$ terms of a non-constant $A.P., a_1, a_2, a_3, ……$ be $50\,n\, + \,\frac{{n\,(n\, - 7)}}{2}A,$ 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

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