JEE Mains · Maths · STD 11 - 8. sequence and series
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
- A \((A, 50 + 46A)\)
- B \((A, 50 + 45A)\)
- C \((50, 50 + 45A)\)
- D \((50, 50 + 46A)\)
Answer & Solution
Correct Answer
(A) \((A, 50 + 46A)\)
Step-by-step Solution
Detailed explanation
\({S_n} = 50n + \frac{{n\left( {n - 7} \right)}}{2}A\) \({T_n} = {S_n} - {S_{n - 1}}\) \( = 50n + \frac{{n\left( {n - 7} \right)}}{2}A - 50\left( {n - 1} \right) - \frac{{\left( {n - 1} \right)\left( {n - 8} \right)}}{2}A\)…
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