Sample QuestionsPART - 2 CH - 8 Sequences and Series questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
If $G _1$ and $G _2$ are two geometric means between $a$ and $b$, then $G _1 G _2$ is equal to :
- A
$\sqrt{a b}$
- ✓
$a b$
- C
$(a b)^2$
- D
$(a b)^3$
Answer: B.
View full solution →For what value of $n$, the geometric mean of $a$ and $b$ is given by $\frac{a^{n+1}+b^{n+1}}{\left(a^n+b^n\right)}$ ?
Answer: D.
View full solution →If third term of geometric progression is 2 then, product of first five terms is :
Answer: C.
View full solution →The number of terms in progression $96,48,24,12$, ............, $\frac{3}{16}$ is :
Answer: B.
View full solution →The common ratio of geometric progression $\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3 \sqrt{3}}, \ldots \ldots$ is :
- ✓
$\frac{1}{3}$
- B
$\frac{1}{\sqrt{3}}$
- C
$\sqrt{3}$
- D
Answer: A.
View full solution →If $\log _x a, a^{x / 2}$ and $\log _b x$ are in geometric progression, then the value of $x$ is $\log _a\left(\log _b a\right)$.
View full solution →If 5th term of a geometric progression is 2 then sum of its 9 terms is 521 .
If $a, b$ and $c$ are in geometric progression, then $\frac{1}{\log _a m}, \frac{1}{\log _b m}$, and $\frac{1}{\log _c m}$ are in geometric progression.
View full solution →$a_1, a_2, a_3$, __________ is a sequence, then the sum expressed as $a_1+a_2+a_3+$.......... is called ___________
View full solution →The sequences in which the number of terms is limited are called ______________
According to a rule, arrangement of numbers in a defined order is called ______________
The geometric mean of two positive numbers $a$ and $b$ is ______________
View full solution →If first term of G.P. is $a$, last term is $l$ and common ratio is $r$, then find its sum.
View full solution →Insert five GM.s between 3 and 192.
What is the 8th term of the G.P. $2,1, \frac{1}{2}, \frac{1}{4}, \ldots \ldots . ?$
View full solution →Write the first three terms of the sequence whose $n^{\text {th }}$ term is given by $a_n=\frac{n^2}{3^n}$
View full solution →Find the sum of odd numbers from 1 to 2001.
Insert 5 G.M.s between $\frac{1}{3}$ and 9 and verify that their product is the 5 th power of the G.M. between $\frac{1}{3}$ and 9 .
View full solution →Find four numbers forming a geometric progression in which the third term is greater than the first term by 9 and the second term is greater than the 4th by 18.
If $S_1, S_2, S_3$ are respectively the sum of $n, 2 n$ and $3 n$ terms of G.P. then prove that $S_1{ }^2+S_2{ }^2=S_1$ $\left(S_2+S_3\right)$.
View full solution →If $a, b, c$ are in G.P. prove that:
(i) $\log a^n, \log b^n, \log c^n$ are in G.P.
(ii) $\log a, \log b, \log c$ are in G.P.
(iii) $\left(a^2+b^2\right)\left(a^2+c^2\right)=(a b+b c)^2$
(iv) $a\left(b^2+c^2\right)=c\left(a^2+b^2\right)$
View full solution →For a G.P., if $(m+n)^{\text {th }}$ term is $P$ and $(m-n)$ th term is $q$, then prove that $m^{\text {th }}$ and $q^{\text {th }}$ term are $\sqrt{p q}$ and $p\left(\frac{q}{p}\right)^{m / 2 n}$ respectively.
View full solution →| Part (A) | Part (B) |
| 1 The 9th term of geometric progression $1,4,16,64, \ldots$ | (a) $\sqrt{3}\left(\frac{1}{3}\right)^{n-1}$ |
| 2. The 10 th term of geometric progression $-\frac{3}{4}, \frac{1}{2},-\frac{1}{3}, \frac{2}{9}, \ldots \ldots$ is | (b) $4^8$ |
| 3. The $n$th term of geometric progression $\sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3 \sqrt{3}}, \ldots \ldots $ is | (c) 2186 |
| 4. The sum of seven terms of geometric progression $2,6,8$,$\ldots$ is | (d) $\sqrt{7}\left(\frac{3^{n / 2}-1}{\sqrt{3}-1}\right)$ |
| 5. The sum of 10 terms of geometric progression $4,2,1,1 / 2$, $\ldots$ is | (e) $8\left(1-\frac{1}{1024}\right)$ |
| | (f) $\frac{1}{2}\left(\frac{2}{3}\right)^8$ |