Common Ratio

Q.

If in an infinite G.P., first term is equal to 10 times the sum of all successive terms, then is common ratio is

1. $\frac{1}{10}$

2. $\frac{1}{11}$

3. $\frac{1}{9}$

4. $\frac{1}{20}$

Q.

The terms of a G.P. are positive. If each term is equal to the sum of two terms that follow it, then the common ratio is

1. 

2. 

3. 1

Q.

In a sequence of $21$ terms, the first $11$ terms are in AP with common difference $2$ and the last $11$ terms are in GP with common ratio $2$. If the middle term of an AP is equal to the middle term of the GP, then the middle term of the entire sequence is

1. $-\frac{10}{31}$

2. $\frac{10}{31}$

3. $\frac{32}{31}$

4. $-\frac{31}{32}$

Q.

Consider an infinite G.P with first term $a$ and common ratio $r$, its sum is $4$ and the second term is $\frac{3}{4}$, then

1. $a=\frac{4}{7},r=\frac{3}{7}$

2. $a=\frac{3}{2},r=\frac{1}{2}$

3. $a=2,r=\frac{3}{8}$

4. $a=3,r=\frac{1}{4}$

Q.

The sum of the series$\frac{9}{19}+\frac{99}{{19}^2}+\frac{999}{{19}^3}+.............\infty$is

1. $\frac{19}{18}$

2. $\frac{18}{19}$

3. $\frac{7}{18}$

4. None of these

Q.

Find the ninth term of the GP: $1,4,16,64,....$.

Q. If the ratio of two numbers is $5:3$ and their L.C.M. is $75$, then the sum of their squares is

Q.

In a G.P. of even number of terms, the sum of all terms is five times the sum of the odd terms. The common ratio of the G.P. is

1. $-\frac{4}{5}$

2. $\frac{1}{5}$

3. none of these

4. 4

Q. the ratio of the sum of n terms of two APs is (3n+1):(4n+3). find the ratio of their mth terms

Q.

If the ratio of sum of m terms and n terms of an A.P. is , m²:n² then prove that the ratio of its mth and nth terms will be 2m-1 : 2n-1.

I know the answer for this, but i want to confirm if this method is right because everywhere, the other method is given!

S(m)÷s(n) = m²/n²
On solving till a point, we get this equation

2an + mnd -nd = 2am + mnd -dm
→ 2an-nd = 2am-dm
→ n(2a-d) = m(2a-d)
→ n=m. ......(1)

In the 'to prove' statement,

LHS →
a+(m-1)d / a+(n-1)d

Replacing n with m (using 1)

a+(m-1)d/a+(m-1)d = 1/1 = 1


RHS →

2m-1/2n-1
Replacing n with m (using 1)
2m-1/2m-1
= 1/1
=1

Hence, LHS=RHS.




IS THIS CORRECT?

Q. For what values of $x$, the numbers $-\frac{2}{7},x,-\frac{7}{2}$ are in $G.P$ ?

Q.

$a,b,c$ are in G.P. $a+b+c=bx$ , then $x$ cannot be;

1. $2$

2. $-2$

3. $3$

4. $4$

Q. Find out the wrong number in the series given below :
$6,13,27,55,110,223$

1. $27$
2. $55$
3. $110$
4. $223$

Q. For a sequence, if $S_n=3(2^n-1)$, find $n^{th}$ term, hence show that it is a G.P. Also find the common ratio.

Q. The first terms of a G.P. is $1$. The sum of the third and fifth term is $90$. Find the common ratio.

Q.

Find two positive numbers x and y such that x + y = 60 and xy³ is maximum.

Q. The second, the first, and the third term of an arithmetic progression, whose common difference is nonzero, form a geometric progression in that order. Find its common ratio.

Q.

Show that the products of the corresponding terms of the sequences form a G.P, and find the common ratio.

Q. The common ratio of a G.P. is 3 and the last term is 486. If the sum of these terms be 728, find the first term.

Q. The sum of first three terms of a G.P. is $16$ and the sum of the next three terms, is $128$. Determine the first term, the common ratio and the sum to n terms of the G.P.

Q.

If a, b, c are in G.P, then

1. a(b² + c²) = c(b² + c²)
2. a(b² + c²) = c(a² + b²)
3. a²(b + c) = c²(a + b)
4. none of these

Q.

In a sequence of 21 tems, the first 11 terms are in A.P. with common difference 2 and the last 11 terms are in G.P. with common ratio 2. If the middle term of the A.P. is equal to the middle term of the G.P., then the middle term of the entire sequence is

1. -10/31

2. 10/31

3. 32/31

4. -31/32

Q. Show that if f₁ and f₂ are one-one maps from R to R, then the product $f_1\times f_2:R\to R$ defined by $\left(f_1\times f_2\right)\left(x\right)=f_1\left(x\right)f_2\left(x\right)$ need not be one-one.

Q. In a sequence of 4n+1 terms, the first 2n+1 terms are in a.p. whose d is 2 and the last 2n+1 terms are in g.p. whose r is 0.5 , if the middle terms of a.p. and g.p. are equal, then middle term of sequence in terms of only n.

Q.

If the 4^th, 10^th and 16^th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

Q. If $G_1,G_2,G_3....G_n$ are n numbers inserted between numbers $a$ and $b$ of a G.P. then common ratio is

1. $(\frac{a}{b}{)}^{\frac{1}{n+1}}$
2. $(\frac{b}{a}{)}^{\frac{1}{n+1}}$
3. $(\frac{b}{a}{)}^{n+1}$
4. $(\frac{a}{b}{)}^{n+1}$

Q. $-3,-15,-75,-\mathrm{375....}$ is a geometric sequence. Find the common ratio.

1. $3$
2. $-5$
3. $5$
4. $-3$

Q. The sum of three terms of a geometric sequence is $\frac{39}{10}$ and their product is $1$. Find the common ratio and the terms

Q. The common ratio of the GP $3(2{)}^3$, $3(2{)}^4$, $3(2{)}^5$ is

1. 1
2. 2
3. 3
4. 4

Q. The first term of a geometric series is $375$ and the fourth term is $192$. Find the common ratio and the sum of the first $14$ terms