(D) Given the recurrence relation: $a_{n+1} - a_n = n - 2$. Summing this from $n = 1$ to $50$: $\sum_{n=1}^{50} (a_{n+1} - a_n) = \sum_{n=1}^{50} (n -

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(D) Given the recurrence relation: $a_{n+1} - a_n = n - 2$. Summing this from $n = 1$ to $50$: $\sum_{n=1}^{50} (a_{n+1} - a_n) = \sum_{n=1}^{50} (n - 2)$ $a_{51} - a_1 = \sum_{n=1}^{50} n - \sum_{n=1}^{50} 2$ $a_{51} - 5 = \frac{50 \times 51}{2} - (50 \times 2)$ $a_{51} - 5 = 1275 - 100$ $a_{51} - 5 = 1175$ $a_{51} = 1180$.

Class 11 Mathematics Sequences and Series nth term of special series, Sum to n terms and Infinite number of terms

Let $a_n$ be a sequence such that $a_1 = 5$ and $a_{n+1} = a_n + (n - 2)$ for all $n \in N$. Then $a_{51}$ is:

A
$1165$
B
$1170$
C
$1175$
D
$1180$

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Sequences and Series

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nth term of special series, Sum to n terms and Infinite number of terms

Let $S_k = \frac{1 + 2 + 3 + .... + k}{k}$. If $S_1^2 + S_2^2 + ....... + S_{10}^2 = \frac{5}{12}A$,then $A$ is equal to:

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If $S_{1}, S_{2}, S_{3}$ are the sum of first $n$ natural numbers,their squares,and their cubes,respectively,show that $9 S_{2}^{2} = S_{3}(1 + 8 S_{1})$.

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The first term of the ${11^{th}}$ group in the following sequence of groups $(1), (2, 3, 4), (5, 6, 7, 8, 9), \dots$ is

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The product $(32)(32)^{1/6}(32)^{1/36} \dots \infty$ is

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If $S_n = 1^3 + 2^3 + \ldots + n^3$ and $T_n = 1 + 2 + \ldots + n$,then

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Let $a_n$ be a sequence such that $a_1 = 5$ and $a_{n+1} = a_n + (n - 2)$ for all $n \in N$. Then $a_{51}$ is:

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