AP GP HP Questions and Answers

As we know the formula of nth term of GP = ar^n-1
9th term = ar⁸ and 5th term =ar⁴
Given that
ar⁸ = 16ar⁴
=> r⁴ = 16
=> r = 2
Given ar⁶ = 96
put value of r above
a(2)⁶ = 96
a = 96/64 = 1.5
Hence, first term of the GP is 1.5.

Arithmetic mean of 61 means there total is 61 multiplied by 2 = 122
and, Geometric Mean of 11 meaning there product will be square of 11 = 121
Clearly, X = 121

Y = 1

Thus, value of X is 121.

As we know a = 1

d = 2

n = 50 .

Then,

s_{n} =\frac{n}{2}[2a+(n-1)\times d]

S₅₀= 25 × (2 × 1 + (50 − 1) × 2)

= 25 × (2 + 49 × 2)

= 25 × (2 + 98)

= 2500.

l = a + (n − 1)d

=>100 = 1 + (n − 1) × 1·5

Now by subtracting 1 from each side:

99 = (n − 1) × 1·5

On adding 1·5 to both the side, we will get 66 = n − 1

n = 67.

Now by using formula

s_{n} =\frac{n}{2}(a+l)

s_{67} =\frac{67}{2}(1+100)

= 3383.5.

a = first term

d = common difference

Now a = 4

S_{9} = 2\times S_{6}------(i)

Formula:

S_{n} =\frac{n}{2}[2a+(n-1)\times d]

By applying (i)



=>4.5(8+8d)= 6(8+5d)

=>36+36d=48+30d

6d=12

d = 2.

Given AP 6 , 9.5 , 13,......

First Term a = 6

Common difference = 9.5-6 = 3.5

Now, formula is

s_{n} =\frac{n}{2}[2a+(n-1)\times d]

To find the sum to 22 terms of an AP,

n = 22

Thus, S = \frac{22}{2}[2\times 6 + (21)3.5]

=> S = 11[12 + 73.5]

=> S = 11 x 85.5

S = 940.5

Solution:

S_{n} = \frac{n}{2} [2a + (n-1)d]

Consider sum of first c terms,

S_{c} = \frac{c}{2} [2a + (c-1)d]

Consider sum of first e terms,

S_{e} = \frac{e}{2} [2a + (e-1)d]

Given,

S_{c} = S_{e}

\frac{c}{2} [2a + (c-1)d]= \frac{e}{2} [2a + (e-1)d]

c[2a+(c−1)d]=e[2a+(e−1)d]

2ac +cd(c-1) = 2ae +ed(e-1)

2a(c−e) + d[c(c−1)−e(e−1)] = 0

2a(c-e) + d[c^{2}-c-e^{2} + e] = 0

2a(c−e)+d[(c+e)(c−e) - (c-e)] = 0

2a(c−e)+d[(c−e) (c+e-1)] = 0

2a+d(c+e-1) = 0

Now,

S_{c+e} = \frac{c + e}{2}[2a + (c + e- 1)d]

= \frac{c+e}{2} \times 0 = 0

=0

Therefore the sum of c+e terms is 0.

X + Y = 9

x³ + y³ = 243
Let’s assume x = 3 , y = 6

So: 3³+6³=243

x² + y² = 9 + 36 = 45.

S_{48}= 4 \times S_{36}

\frac{48}{2}(2a+47d) = 4\times \frac{36}{2}(2a+35d)

4a+58d=0

or 2a+29d=0

S_{30} = \frac{30}{2}(2a+29d)

=15(0)

=0.

X + Y = 12

x³ + y³ = 243
Let’s assume x = 4 , y = 8

So: 4³+8³=576

x² + y² = 16 + 64 = 80.

If p, q, r pre in h.p then \frac{1}{p},\frac{1}{q},\frac{1}{r} will be in A.P. series and \frac{1}{q} will be the AM of \frac{1}{p},\frac{1}{r}

Let r be the harmonic mean,

r= \frac{2pq}{(p+q)}

Given the sequence t_{1}=7 , t_{n}= −3t_{n-1} we have

t₁=7

t₂=−3 t_{1}=−21

t₃=−3t₂ = 63

t₄=−3t₃=−189

t₅=−3t₄= 567

t₆=-3t₅= -1701

As we know: in geometric mean problems ab=√ab

Here A=20, B=25

G.M. of ab=√ab

Ab=√20x25

Ab=√500

Ab=22.56

According to the n−th term formula we can write, that:

a₉ = a_{1}\times r^{8}, thus

r^{8} = \frac{a_{9}}{a_{8}}

r = (\frac{a_{9}}{a_{8}})^{\frac{1}{8}}

The nth term of a geometric series is

a_{n} = a\times r^{n-1}

a = first term

r = common ratio

r =\frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}} =......=\frac{a_{n}}{a_{n-1}}

Now a=4 and r = \frac{12}{4}= 3

so=a₉= 4x3⁸ = 26244.

A.M.= X+Y/2

G.M=√XY

H.M=2XY/X+Y

Now we have:

√G.M=22

G.M=22²

G.M=22X22=484

A.M=26

Or x+y/2=26

x+y=52

Now H.M=2XY/X+Y

Or H.M= 2X484/52

H.M=18.61.

As we know: Arithmetic Mean* Harmonic Mean= Geometric Mean²

1 x Harmonic Mean= Geometric Mean²

Or Harmonic Mean= Geometric Mean²

So it is true.

Let’s assume two number as x and y.

Now A.M.

=>\frac{x+y}{2} = 28

x+y = 56

HM = \frac{2xy}{(x+y)}

So x=15 and y=31

H.M= \frac{(2\times 15 \times 31)}{(15+31)}

H M = \frac{930}{46} = 20.21.

The mean of the reciprocals of the terms = 1/100 and 1/69.

Thus, the total of the reciprocals of the terms in set i=

1/100×25=1/4
and in set ii=1/69×23=1/3

The sum of the reciprocals of all 48 components is 1/4+1/3=7/12.

Hence their mean is 7/12÷48=82.28

H.M = \frac{26^{2}}{30} = 22.53

For number 15,12,18,20 harmonic mean is -
\frac{4}{( \frac{1}{15} + \frac{1}{12} + \frac{1}{18} +\frac{1}{20})}
= \frac{4}{0.06+0.08+0.05+0.05}
harmonic mean = 16.66

: the above table represents the grouped data in which age is variable and number of scholars shows frequency.

Assume x and y be the number.

A.M.=X+Y/2

G.M=√XY

H.M.= 2XY/X+Y

A.M x  H.M= X+Y/2 * 2XY/X+Y

XY=(G.M)²

A.M x H.M=(G.M)²

18 x H.M=(15)²

H.M= \frac{225}{18} = 12.5

HM = \frac{36^{2}}{40} = 32.4

We have: {5, 9, 7, 10, 8, 6}

H.M = \frac{6}{\frac{1}{5} +\frac{1}{9}+\frac{1}{7}+\frac{1}{10}+\frac{1}{8} +\frac{1}{5}}

= \frac{6}{0.2+0.11+0.14+0.1+0.12+0.2} =6.89

 

 

A.M.= Sum of the numbers = 26.4 + 29.8 + 25.8 = 82

Arithmetic Mean = 82 ÷ 3 = 27.3

Harmonic Mean =Sum of the reciprocal = (1 ÷ 26.4) + (1 ÷ 29.8) + (1 ÷ 25.8)  =

0.037+0.033+0.038=0.108

Harmonic Mean = 3 ÷ 0.108 = 27.77

Nth term of H.P

T_{n} = \frac{1}{a + (n-1)d}

Which means nth term of H.P is \frac{1}{\text{ nth term of A.P}}

Nth term of A.P = a + (n-1)d

where,

n = 9

a = 4

d = 4

T_{n} = 4 + (9-1)4

T_{9} = 4 + 32

T_{9} = 36

HP_{9} = \frac{1}{A.P}

HP_{9} = \frac{1}{36}

 

If P, Q, R are in hp then P^-1, Q^-1, R^-1 is in AP