The 99^{th} term of the series 2 + 7 + 14 + 2 | vedclass

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(A) Let the $n^{th}$ term be $T_n$. The series is $2, 7, 14, 23, 34, \dots$The differences between consecutive terms are $5, 7, 9, 11, \dots$,which fo

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$9998$

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(A) Let the $n^{th}$ term be $T_n$. The series is $2, 7, 14, 23, 34, \dots$The differences between consecutive terms are $5, 7, 9, 11, \dots$,which form an Arithmetic Progression $(AP)$.Let $T_n = an^2 + bn + c$.For $n=1, T_1 = a + b + c = 2$For $n=2, T_2 = 4a + 2b + c = 7$For $n=3, T_3 = 9a + 3b + c = 14$Subtracting equations: $(4a+2b+c) - (a+b+c) = 3a+b = 5$ and $(9a+3b+c) - (4a+2b+c) = 5a+b = 7$.Subtracting these results: $2a = 2 \Rightarrow a = 1$.Substituting $a=1$ into $3a+b=5$,we get $3+b=5 \Rightarrow b=2$.Substituting $a=1, b=2$ into $a+b+c=2$,we get $1+2+c=2 \Rightarrow c=-1$.Thus,$T_n = n^2 + 2n - 1$.For $n=99$,$T_{99} = (99)^2 + 2(99) - 1 = 9801 + 198 - 1 = 9998$.

The $99^{th}$ term of the series $2 + 7 + 14 + 23 + 34 + \dots$ is

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