If ${a_1} = 1$ and ${a_{n + 1}} = \frac{{4 + 3{a_n}}}{{3 + 2{a_n}}},\;n \ge 1$ and if $\mathop {\lim }\limits_{n \to \infty } {a_n} = a$, then the value of $a$ is

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

{a_1} = 1

and

{a_{n + 1}} = \frac{{4 + 3{a_n}}}{{3 + 2{a_n}}},\;n \ge 1

and if

\lim\limits_{n \to \infty } {a_n} = a

, then the value of

a

View answer and explanation

Correct option: A

\sqrt 2

Option A (

\sqrt 2

) is correct because the limit of the sequence is found by setting

a_{n+1} = a_n = a

and solving the resulting quadratic equation, discarding the negative root since all terms are positive.

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