Let aₙ be a sequence such that a₁ = 5 and a_{n+1} = aₙ + (n -2) for all n ∈ N , then a

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8. Sequences and Series

normal

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

$1165$

$1170$

$1175$

$1180$

Solution

$\begin{array}{*{20}{c}}
{{a_{n + 1}} – {a_n} = n – 2}\\
{{a_{51}} – {a_{50}} = 48}\\
{{a_{50}} – {a_{49}} = 47}\\
\vdots \\
{{a_3} – {a_2} = 0}\\
{{a_2} – {a_1} = – 1}
\end{array}$

${a_{51}} – {a_1} = – 1 + 0 + 1 + 2 + …… + 48$

${a_{51}} – 5 = 1175 \Rightarrow {a_{51}} = 1180$

Standard 11

Mathematics

Let $a_{1}, a_{2} \ldots, a_{n}$ be a given $A.P.$ whose common difference is an integer and $S _{ n }= a _{1}+ a _{2}+\ldots+ a _{ n }$ If $a_{1}=1, a_{n}=300$ and $15 \leq n \leq 50,$ then the ordered pair $\left( S _{ n -4}, a _{ n -4}\right)$ is equal to

hard

(JEE MAIN-2020)

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hard

(JEE MAIN-2022)

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hard

(JEE MAIN-2023)

The sums of $n$ terms of three $A.P.'s$ whose first term is $1$ and common differences are $1, 2, 3$ are ${S_1},\;{S_2},\;{S_3}$ respectively. The true relation is

hard

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hard

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

Solution

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