Sum of first five terms of series 3 + 4frac{1}{2} + 6frac{3}{4} + ...... will be

English

Gujarati

Hindi

8. Sequences and Series

easy

The sum of the first five terms of the series $3 + 4\frac{1}{2} + 6\frac{3}{4} + ......$ will be

$39\frac{9}{{16}}$

$18\frac{3}{{16}}$

$39\frac{7}{{16}}$

$13\frac{9}{{16}}$

Solution

(a) Given series is $3 + 4\frac{1}{2} + 6\frac{3}{4} + …….. = 3 + \frac{9}{2} + \frac{{27}}{4} + …..$

$ = 3 + \frac{{{3^2}}}{2} + \frac{{{3^3}}}{4} + \frac{{{3^4}}}{8} + \frac{{{3^5}}}{{16}} + …..$(in $G.P.$)

Here $a = 3,\;r = \frac{3}{2}$ , then sum of the five terms

${S_5} = \frac{{a({r^n} – 1)}}{{r – 1}} = \frac{{3\left[ {{{\left( {\frac{3}{2}} \right)}^5} – 1} \right]}}{{\frac{3}{2} – 1}} = \frac{{1\left[ {\frac{{{3^5}}}{{32}} – 1} \right]}}{{\frac{1}{2}}}$

$ = 6\left[ {\frac{{243 – 32}}{{32}}} \right] = \frac{{211 \times 3}}{{16}} = \frac{{633}}{{16}} = 39\frac{9}{{16}}$.

Standard 11

Mathematics

Let $\left\langle a_n\right\rangle$ be a sequence such that $a_0=0, a_1=\frac{1}{2}$ and $2 a_{n+2}=5 a_{n+1}-3 a_n, n=0,1,2,3, \ldots \ldots$. Then $\sum_{k=1}^{100} a_k$ is equal to :

hard

(JEE MAIN-2025)

The value of ${4^{1/3}}{.4^{1/9}}{.4^{1/27}}………..\infty $ is

easy

Let $P(x)=1+x+x^2+x^3+x^4+x^5$. What is the remainder when $P\left(x^{12}\right)$ is divided by $P(x)$ ?

normal

(KVPY-2009)

For $0<\mathrm{c}<\mathrm{b}<\mathrm{a}$, let $(\mathrm{a}+\mathrm{b}-2 \mathrm{c}) \mathrm{x}^2+(\mathrm{b}+\mathrm{c}-2 \mathrm{a}) \mathrm{x}$ $+(c+a-2 b)=0$ and $\alpha \neq 1$ be one of its root. Then, among the two statements

$(I)$ If $\alpha \in(-1,0)$, then $\mathrm{b}$ cannot be the geometric mean of $\mathrm{a}$ and $\mathrm{c}$

$(II)$ If $\alpha \in(0,1)$, then $\mathrm{b}$ may be the geometric mean of $a$ and $c$

hard

(JEE MAIN-2024)

The sum of some terms of $G.P.$ is $315$ whose first term and the common ratio are $5$ and $2,$ respectively. Find the last term and the number of terms.

medium