\(\sum_{n=2}^5 (n^2- 3)\)

Putting n = 2, 3, 4, 5

= (2² - 3) + (3² - 3) + (4² - 3) + (5² - 3)

= (4 - 3) + (9 - 3) + (16 - 3) + (25 - 3)

= 1 + 6 +13 +22

= 42 _Ans

u_{n + 1} = 1 - \(\frac {1}{u_n}\) and u₁ = 3, u₂ and u₃ = ?

If n = 1,u_{n + 1} = 1 - \(\frac {1}{u_n}\)

u_{1 + 1} =1 - \(\frac {1}{u_1}\)

or, u₂ = 1 - \(\frac 13\) = \(\frac {3 - 1}{3}\) = \(\frac 23\)

If n = 2, u_{n + 1} =1 - \(\frac {1}{u_n}\)

u2 + 1 =1 - \(\frac {1}{u_2}\)

or, u₃ = 1 - \(\cfrac{1}{\cfrac{2}{3}}\) = 1 - \(\frac 32\) = \(\frac {2 - 3}{2}\) = -\(\frac 12\)

∴ u₂ = \(\frac 23\) and u₃ = -\(\frac 12\) _Ans

Sequence: A sequence is a set of numbers or quantities, which are formed according to some governed laws.

i.e. 1, 4, 16 ............................ and 2, 3, 4, 5...................

Series: The sum of the term of a sequence is called a series.

i.e. 1 + 4 + 16 + ....................... and 2 + 3 + 4 + 5 + ......................

Here,

\(\sum_{n=4}^7 (3n- 2)\)

where: n = 4, 5, 6, 7

\(\sum_{n=4}^7 (3n- 2)\)

= (3×4-2) + (3×5-2) +(3×6-2) +(3×7-2)

= (12-2) + (15-2) + (18-2) + (21-2)

= 10 + 13 + 16 + 19

= 58 _Ans

\(\sum_{k=3}^7 (k^2 + 1)\)

= (3² + 1) +(4² + 1) +(5² + 1) +(6² + 1) +(7² + 1)

= (9 + 1) +(16 + 1) +(25 + 1) +(36 + 1) +(49 + 1)

= 10 + 17 + 26 + 37 + 50

= 140 _Ans

Here,

First term (a) = 1

Second term (b) = 4

Common Difference (d) = b - a = 4 - 1 = 3

Last term (l) = 34

We know,

last term (l) = a + (n - 1)d

or, 34 = 1 + (n - 1) 3

or, 34 = 1 + 3n - 3

or, 3n - 2 = 34

or, 3n = 34 + 2

or, n = \(\frac {36}{3}\)

∴ n = 12

Again,

S_n = \(\frac n2\) [a + l]

S_n = \(\frac{12}{2}\) [1 + 34] = 6× 35 = 210

∴ S_n = 210 _Ans

Here,

first term (a) = 2

second term (b) = 7

common difference (d) = b - a = 7 - 2 = 5

number of items (n) = 16

sum (S₁₆) = ?

We know that,

S_n = \(\frac n2\) [2a + (n - 1) d]

S₁₆ = \(\frac {16}{2}\) [2 × 2 + (16 - 1) 5] = 8 [4 + 15× 5] = 8 [4 + 75] = 8 × 79 = 632 _Ans

Here,

first term (a) =3\(\frac12\) = \(\frac 72\)

second terrn (b) = 1

common difference (d) = b - a = 1 - \(\frac 72\) = \(\frac {2 - 7}{2}\) = -\(\frac 52\)

no. of items (n) = ?

last term (l) =-21\(\frac 12\) = -\(\frac {43}{2}\)

We know that,

l = a + (n - 1) d

or,-\(\frac {43}{2}\) = \(\frac 72\) + (n - 1) (-\(\frac 52\))

or, - 43 = 7 - 5n + 5

or, 5n = 12 + 43

or, n = \(\frac {55}{5}\) = 11

∴-21\(\frac 12\) is the 11^th term. _Ans

Here,

first term (a) = 20

common difference (d) = -3

numbers of term (n) = 7

seventh term (t₇) = ?

We know that,

t_n= a + (n - 1) d

t₇ = 20 + (7 - 1) (-3) = 20 + 6 (-3) = 20 - 18 = 2 _Ans

Here,

first term (a) = 2

second term (b) = 4

common difference (d) = b - a = 4 - 2 = 2

no. of terms (n) = 20

sum (S₂₀) = ?

We know that,

S_n = \(\frac n2\) [2a + (n - 1) d]

S₂₀ = \(\frac {20}{2}\) [2× 2 + (20 - 1)× 2] = 10 [4 + 38] = 10× 42 = 420 _Ans

Here,

first term (a) = 1

second term (b) = 3

common difference (d) = b - a = 3 - 1 = 2

number of items (n) = 20

twenty term (t₂₀) = ?

We know,

t_n= a + (n - 1) d

t₂₀ = 1 + (20 - 1) 2

t₂₀ = 1 + 19× 2 = 1 + 38 = 39 _Ans

Here,

first term (a) = 25

second term (b) = \(\frac {45}{2}\)

common difference (d) = b - a = \(\frac {45}{2}\) - 25 = \(\frac {45 - 50}{2}\) = -\(\frac 52\)

last term (l) = - 15

number of terms (n) = ?

We know that,

l = a + (n - 1) d

or, - 15 = 25 + (n - 1) - \(\frac 52\)

or, - 15 - 25 = -\(\frac 52\)n + \(\frac 52\)

or, - 40 - \(\frac 52\) = -\(\frac 52\)n

or - \(\frac {80 - 5}{2}\) = -\(\frac {5n}{2}\)

or, -\(\frac {85}{2}\)× -\(\frac {2}{5}\) = n

∴ n = 17

∴ The number of term = 17. _Ans

Here,

first term (a) = 3

last term (l) = -9

number of terms (n) = 4 [including a and b]

common difference (d) = \(\frac {b - a}{n - 1}\) = \(\frac {-9 - 3}{4 - 1}\) = -\(\frac {12}{3}\) = - 4

x = a + d = 3 - 4 = - 1

y = a + 2d = 3 + 2× -4 = 3 - 8 = - 5

∴ x = -1 and y = -5 _Ans

Here,

first term (a) = 2

second term (b) = - 9

common difference (d) = b - a = -9 - 2 = -11

last term (l) = -130

number of terms (n) = ?

We know,

l = a + (n - 1) d

or, -130 = 2 + (n - 1) (-11)

or, -130 - 2 = - 11n + 11

or, -132 - 11 = -11n

or, -11n = -143

or, n = \(\frac {-143}{-11}\)

∴ n = 13

Again, Sn = \(\frac n2\) (a + 1) = \(\frac {13}{2}\) (2 - 130) = 6.5× -128

∴ Sn = -832 _Ans

Here,

common difference (d) = -3

number of terms (n) = 7

sum (S₇) = 0

first term (a) = ?

We know,

Sn = \(\frac n2\) [2a + (n - 1) d]

or, 0 = \(\frac 72\) [2a + (7 - 1) (-3)]

or, 0× \(\frac 27\) = 2a + 6× (-3)

or, 0 = 2a - 18

or, 2a = 18

or, a = \(\frac {18}{2}\) = 9

∴ first term (a) = 9 _Ans

Here,

first term (a) = \(\frac 14\)

second term (b) = \(\frac 12\)

common ratio (r) = \(\frac ba\) =\(\cfrac{\frac{1}{2}}{\cfrac{1}{4}}\) = \(\frac 12\)× \(\frac 41\) = 2

last term (l) = 128

number of terms (n) = ?

We know that,

l = ar^n-1

or, 128 = \(\frac 14\) (2)n-1

or, 2⁷× 2² = 2^n-1

or, 2^n-1 = 2⁹

or, n -1 = 9

or, n = 9 + 1

∴ n = 10

∴ The number of term = 10 _Ans

Here,

first term (a) = 16

last term (l) = 36

We know that,

geometric mean (G.M.) =\(\sqrt {ab}\)

G.M. =\(\sqrt{({16}×{36})}\)

= \(\sqrt {({4^2}×{6^2})}\)

= \(\sqrt {({4×6})^2}\)

= \(\sqrt {({24})^2}\)

∴ Geometric mean = 24 _Ans

Here,

first term (a) = \(\frac 13\)

second term (b) = 1

common ratio (r) = \(\frac ba\) = \(\cfrac {1}{\cfrac 13}\) = 3number of terms (n) = 8

number of terms (n) = 8

eighth term (t₈) = ?

Using formula,

t_n = ar^n-1

t₈ = (\(\frac 13\)) (3)^8-1 = \(\frac 13\)× 3× 3⁶ = 729 _Ans

Here,

first term (a) = 3

second term (b) = 1

common ratio (r) = \(\frac ba\) = -\(\frac 13\)

last term (l) = -\(\frac {1}{81}\)

number of terms (n) = ?

We know that,

l = ar^n-1

or, -\(\frac {1}{81}\) = 3 (-\(\frac 13 \))^n-1

or, - \(\frac {1}{3^4× 3}\) = (-\(\frac 13\))^n-1

or, (-\(\frac 13\))⁵ =(-\(\frac 13\))^n-1

or, n - 1 = 5

or, n = 5 + 1

∴ n = 6

∴ The number of term = 6 _Ans

Here,

first term (a) = \(\frac 23\)

second term (b) = -\(\frac 16\)

common ratio (r) =\(\frac ba\) = \(\cfrac{\frac{-1}{6}}{\cfrac{2}{3}}\)= -\(\frac 16\) × \(\frac 32\) = -\(\frac 14\)

number of items (n) = 5

5^th term (t₅) = ?

We know that,

t_n= ar^n-1

or, t₅ = \(\frac 23\)(-\(\frac 14\))^5-1

or, t₅ = \(\frac 23\)×(-\(\frac 14\))⁴

or, t₅ = \(\frac 23\)× \(\frac {1}{16}\)×\(\frac {1}{16}\) = \(\frac {1}{384}\)

∴ fifth term = \(\frac {1}{384}\) _Ans

Here,

first term (a) = \(\frac 23\)

second term (b) = x

third term (c) = \(\frac 32\)

We know that,

G.M. = \(\sqrt {ab}\)

x = \(\sqrt {(\frac 23) × (\frac 32)}\)

x = 1

r = \(\frac ba\) =\(\cfrac{1}{\cfrac{2}{3}}\) = \(\frac 32\)

y = ar = \(\frac 32\)× \(\frac 32\) = \(\frac 94\)

∴ x = 1 and y = \(\frac 94\) _Ans

Here,

first term (a) = 12

second term (b) = 4 common ratio (r) = \(\frac ba\) = \(\frac 42\) = 2

number of terms (n) = 10

sum of the ten terms (S₁₀) = ?

We know that,

S_n = \(\frac {a(r^n - 1)}{r - 1}\) = \(\frac {2(2^10 - 1)}{2 - 1}\) =\(\frac {2(1024 - 1)}{1}\) = 2× 1023 = 2046

∴ The sum of tenth terms = 2046 _Ans

Here,

first term (a) = 81

second term (b) = 27

common ratio (r) = \(\frac ba\) = \(\frac {27}{81}\) = \(\frac 13\)

number of terms (n) = 5

sum of five terms (S₅) = ?

We know,

S_n = \(\frac {a(r^n - 1)}{r - 1}\)

S₅ = \(\frac {81[(\frac 13)^5 - 1]}{\frac 13 - 1}\)

=\(\frac {81[(\frac {1}{243}) - 1]}{\frac {1 - 3}{3}}\)

=\(\frac {81 × (\frac {1 - 243}{243})}{\frac {-2}{3}}\)

= 81× -\(\frac {242}{243}\)× -\(\frac 32\)

= 121

∴ The sum of five terms = 121 _Ans

Let,

first term = a

common ratio = r

From equation,

S₆ = 9 S₃

or,\(\frac {a(r^6 - 1)}{r - 1}\) = 9 [\(\frac {a(r^3 - 1)}{r - 1}\)] [\(\because\)S_n =\(\frac {a(r^n - 1)}{r - 1}\)]

or, \(\frac {a[(r^3)^2 - (1)^2]}{r - 1}\)×\(\frac {r - 1}{a(r^3 - 1)}\) = 9

or, \(\frac{(r^3 + 1)(r^3 - 1)}{r^3 - 1}\) = 9

or, r³ = 9 - 1

or, r³ = 8

or, r³ = 2³

∴ r = 2

∴ The common ratio = 2 _Ans

Let: the three terms of an G.P. are \(\frac ar\), a and ar

From Question,

\(\frac ar\)× a× r = 729

or, a³ = 9³

∴ a = 9

∴ The second term = 9 _Ans

Here,

first term (a) = 6

third term (b) = x

A.M. = 30

A.M. = \(\frac {a + b}{2}\)

or, 30 = \(\frac {6 + x}{2}\)

or, 60 - 6 = x

∴ x = 54

G.M. = \(\sqrt {ab}\)

G.M. = \(\sqrt {6 × 54}\)

G.M. = \(\sqrt {2^3 × 3^3 ×3^3}\)

G.M. = 18

∴ The G.M. = 18 _Ans

Here,

first term (a) = \(\frac 19\)

last term (l) = 9

number of Geometric mean (n) = 3

total number of term (n) = 5

3 Geometric mean (m₁), (m₂), (m₃) = ?

l = ar^n-1

or, 9 = \(\frac 19\)r^5-1

or, 81 = r⁴

or, 3⁴ = r⁴

∴ r = 3

m₁ = ar = \(\frac 19\)× 3 = \(\frac 13\)

m₂ = ar² = \(\frac 19\)× 32 = \(\frac 19\)× 9 = 1

m₃ = ar³ = \(\frac 19\)× 33 = \(\frac 19\)× 27 = 3

∴ m₁ = \(\frac 13\), m₂ = 1, m₃ = 3 _Ans

If M and G be the arithmetic mean and geometric mean between two positive numbers a and b, then

m = \(\frac {a + b}{2}\) .................... (1)

G = \(\sqrt {ab}\) ................................(2)

Equation (1) - (2) we get

M - G = \(\frac {a + b}{2}\) - \(\sqrt {ab}\)

= \(\frac {a + b - 2\sqrt{ab}}{2}\)

= \(\frac {(\sqrt a)^2 - 2\sqrt a ⋅ \sqrt b + (\sqrt b)^2}{2}\)

= \(\frac 12\) (\(\sqrt a\) - \(\sqrt b\))²

\(\sqrt a\) - \(\sqrt b\) gives positive value

∴ M - G = \(\frac {(\sqrt a - \sqrt b)^2}{2}\)≥ 0

M - G≥ 0 or, M ≥G

∴ A.M.≥ G.M. _Ans

Let:

S_n = 1, 3, 5, ..................... n

First term (a) = 1

Second term (b) = 3

Common difference (d) = b - a = 3 - 1 =2

number of items = n

We know,

S_n = \(\frac n2\)[2a + (n - 1) d]

or, S_n = \(\frac n2\)[ 2× 1 + (n - 1) 2]

or, S_n = \(\frac n2\)[2 + 2n - 2]

or, S_n = \(\frac n2\)× 2n

∴ S_n = n²_Ans

Here,

first term (a) = 1

common ratio (r) = 2

sum (S_n) = 255

We know,S

S_n = \(\frac{a(r^n - 1)}{r - 1}\)

or, S_n =\(\frac{1(2^n - 1)}{2 - 1}\)

or,\(\frac{2^n - 1}{1}\) = 255

or, 2ⁿ = 255 + 1

or, 2ⁿ = 256

or, 2ⁿ = 2⁸

∴ n = 8

∴ The no. of terms = 8 _Ans

Let:

first term (a) = 3

last term (b) = 243

no. of geometric means (n) = 3

common ratio (r) = ?

3 G.M are X₁, X₂, X₃ = ?

r = (\(\frac ba\))^{\(\frac {1}{n+1}\)}= (\(\frac {243}{3}\))^{\(\frac {1}{3+1}\)}= (81)^{\(\frac 14\)}= (3)^{\(\frac {4×1}{4}\)} = 3

X₁ = ar = 3× 3 = 9

X₂ = ar² =3× (3)² = 27

x₃ = ar³ =3× (3)³ = 81

∴ 9, 27, 81 _Ans

Let:

first term (a) = 27

last term (l) = \(\frac {32}{9}\)

number of terms (n) = 4

common ratio (r) = (\(\frac ba\))^{\(\frac {1}{n+1}\)}

= (\(\cfrac{\frac{32}{9}}{27}\))^{\(\frac {1}{4+1}\)}

=( \(\frac {32}{9×27}\))^{\(\frac 15\)}

= (\(\frac 23\))^{\(\frac {5×1}{5}\)}

=\(\frac 23\)

p = ar = 27×\(\frac 23\) = 18

q = ar²=27×\(\frac 23\)×\(\frac 23\) = 12

r = ar³ =27×\(\frac 23\)×\(\frac 23\)×\(\frac 23\) = 8

s = ar⁴ =27×\(\frac 23\)×\(\frac 23\)×\(\frac 23\)×\(\frac 23\) = \(\frac {16}{3}\)

∴ p = 18, q = 12, r = 8 and s = \(\frac {16}{3}\) _Ans

Here,

first term (a) = 2

second term (b) = -6

common ratio (r) = \(\frac ba\) = \(\frac {-6}{2}\) = -3

number of terms (n) = 6

We know that,

S_n = \(\frac {a(1 - r^n)}{1 - r}\)

S_n = \(\frac {2(1 - (-3)^6)}{1 - (-3)}\)

S_n = \(\frac {2(1 -729)}{1 + 3}\)

S_n = \(\frac {2× (-729))}{4}\)

∴S_n= -364.5_Ans

Here,

first term (a) = 3

second term (b) = 6

common ratio (r) = \(\frac ba\) =\(\frac 63\) = 2

last term (l) = 1535

sum of the series (S_n) = ?

Using formula,

S_n = \(\frac{lr - a}{r - 1}\) = \(\frac {1535 × 2 - 3}{2 - 1}\) = \(\frac {3070 - 3}{1}\) = 3067

∴ Sum (S_n) = 3067 _Ans

Let:

first term = a

common difference = d

number of terms (n) = 4

sum of 4 terms (S₄) = 26

We know that,

S_n = \(\frac n2\) [2a + (n - 1) d]

or, S₄= \(\frac 42\) [2a + (4 - 1) d]

or, 26 = 2[2a + 3d]

or, 2a + 3d = 13 ........................(1)

Again,

number of terms (n) = 8

sum of 8 terms (S₈) = 100

S_n = \(\frac n2\) [2a + (n - 1) d]

or, S₈= \(\frac 82\) [2a + (8 - 1) d]

or, 100 = 4[2a + 7d]

or, 2a + 7d = 25 ........................(2)

Subtracting the equation (1) from equation (2)

or, d = \(\frac {12}{4}\)

∴ d = 3

Substituting the value of d in equation (2)

2a + 7d = 25

or, 2a + 7× 3 = 25

or, 2a = 25 - 21

or, a = \(\frac 42\)

∴ a = 2

Sum of 28 terms (S₂₈) = ?

S_n = \(\frac n2\) [2a + (n - 1) d]
or,S28 = \(\frac {28}{2}\) [2(2) + (28 - 1) d]

or,S28 = 14[4 + 27× 3]

or,S28 = 14 (4 + 81)

or,S28 = 14× 85

∴S28 = 1190 _Ans

Here,

first term (a) = 27

second term (b) = 24

common difference (d) = b - a = 24 - 27 = -3

Sum of the series (S_n) = 132

We know that,

S_n = \(\frac n2\) [2a + (n - 1) d]

or, 132 = \(\frac n2\) [2×27 + (n - 1) (-3)]

or, 264 = n[52 - 3n + 3]

or, 264 = n(57 - 3n)

or, 264 = 57n - 3n2

or, 3n² - 57n + 264 = 0

or, 3(n² - 19n + 88) = 0

or, n² - 11n - 8n + 88 = 0

or, n(n - 11) -8(n - 11) = 0

or, (n - 11) (n - 8) = 0

Either, n - 11 = 0 ∴ n = 11

Or, n - 8 = 0 ∴ n = 8

∴n = 11 or 8

∴ The double answer of n show that there are two series in A.S. having the same sum.

The series to 8 terms 27 + 24 + 21 + 18 + 15 + 12 + 9 + 6 and the series to 11 terms is27 + 24 + 21 + 18 + 15 + 12 + 9 + 6 + 3 + 0 - 3

∴ S₈ = S₁₁ =132 _Ans

Here,

third term (t₃₎ = 40

number of terms (n) = 3

Using formula,

t_n= a + (n - 1)d

t₃ = a + (3 - 1)d

40 = a + 2d ............................(1)

Again,

thirteen term (t₁₃) = 0

number of terms (n) = 13

Using formula,

t_n= a + (n - 1)d

t₁₃ = a + (13 - 1)d

0 = a + 12d ...........................(2)

Subtracting equation (2) from (1)

∴ d = -4

Substituting the value of d in equation (2)

a + 12d = 0

a + 12× (-4) = 0

a - 48 = 0

∴ a = 48

Now,

a = 48

d = -4

n = 25

t₂₅ = ?

We know,

t_n= a + (n - 1)d

∴ t₂₅ = 48 + (25 - 1)(-4) = 48 + 24 (-4) = 48 - 96 = - 48 _Ans

Let:

first term = a - d

second term = a

third term = a + d

common difference = d

Let: the threeterms of A.S. are a-d, a, a+d

From first condition.

a - d + a + a+ d = 21........................(i)

or, 3a = 21

or, a = \(\frac {21}{3}\)

∴ a = 7

From second condition,

a + d - (a + a - d) = 9

or, a + d - 2a + d = 9

or, 2d - a = 9.......................... (ii)

Putting the value of a in equation (ii)

2d - 7 = 9

or, 2d = 9 + 7

or, d = \(\frac {16}{2}\) = 8

∴ first term = a - d = 7 - 8 = -1

second term = a = 7

third term = a + d = 7 + 8 = 15

∴Required termsare -1, 7, 15 _Ans

Here,

fourth term (t₄) = 1

no. of terms (n) = 4

t_n= a + (n -1)d

t₄ = a + (4 - 1)d

1 = a + 3d

a = 1 - 3d ..................(1)

sum of eight terms (S₈) = 18

no. of terms (n) = 8

S_n = \(\frac n2\) [2a + (n - 1) d]

Putting the value of a from equation (1),

or, 18 = \(\frac 82\) [2(1 - 3d) + (8 - 1) d]

or, 18 = 4 (2 - 6d + 7d)

or, 18 = 4 (2 + d)

or, 18 = 8 + 4d

or, 4d = 18 - 8

or, d = \(\frac {10}{4}\)

∴ d = \(\frac 52\)

Putting the value of d in equation (1)

a = 1 - 3d = 1 - 3× \(\frac 52\) = 1 - \(\frac {15}{2}\) = \(\frac {2 - 15}{2}\) = -\(\frac {13}{2}\)

Again,

a = -\(\frac {13}{2}\)

d = \(\frac 52\)

n = 10

t₁₀ = ?

We know,

t_n = a + (n - 1)d

t₁₀ = -\(\frac {13}{2}\) + (10 - 1)× \(\frac 52\)

= \(\frac {-13}{2}\) + \(\frac {45}{2}\)

= \(\frac {-13 + 45}{2}\)

= \(\frac {32}{2}\)

= 16

∴ 10^th term = 16 _Ans

Let:

first term = a

fifth term (t₅) = 17

number of terms (n) = 5

common difference = d

We know,

t_n= a + (n - 1)d

or, t₅ = a + (5 - 1)d

or, 17 = a + 4d ....................(1)

Again,

10^th term (t₁₀) = 42

number of terms (n) = 10

We know,

t_n= a + (n - 1)d

or, t₁₀ = a + (10 - 1)d

or, 42 = a + 9d .......................(2)

Subtracting equation (1) and (2)

or, d = \(\frac {-25}{-5}\)

∴ d = 5

Putting the value of d in equation (1)

a + 4d = 17

or, a + 4×5 = 17

or, a = 17 - 20

∴ a = -3

Again,

a = -3

d = 5

n = 20

t₂₀ = ?

We know,

t_n= a + (n - 1)d

t₂₀ = -3 + (20 - 1)5 = -3 + 19× 5 = -3 + 95 = 92

∴ 20th term (t₂₀) = 92 _Ans

Let;

first term (a) =

common difference (d) = ?

number of terms (n) = 10

sum of the 10 terms (S₁₀) = 520

We know that,

S_n = \(\frac n2\) [2a + (n - 1) d]

or, S₁₀= \(\frac {10}{2}\) [2a + (10 - 1) d]

or, 520 = 5[2a + 9d]

or, 2a + 9d = \(\frac {520}{5}\)

or, 2a + 9d = 104 ..........................(1)

Again,

t₇ = 2 t₃

In L.H.S. n = 7 and R.H.S. n = 3

We know, t_n= a + (n - 1)d

a + (n - 1)d = 2 [a + (n - 1)d]

or, a + (7 - 1)d = 2 [a + (3 - 1)d]

or, a + 6d = 2(a + 2d)

or, a + 6d = 2a + 4d

or, 2a - a = 6d - 4d

or, a = 2d ...............................(2)

Putting the value of a in equation (1)

2× 2d + 9d= 104

or, 4d + 9d = 104

or, 13d = 104

or, d = \(\frac {104}{13}\)

∴ d = 8

Putting the value of d in equation (2)

a = 2d = 2× 8 = 16

∴ first term = 16

∴ common difference = 8 _Ans

Here,

first term (a) = 10 years

common difference (d) = 4 months

\(\frac {4}{12}\) years = \(\frac 13\) years

sum of the series = 200 years

number of students (n) = ?

We know,

S_n = \(\frac n2\) [2a + (n - 1) d]

or, 200 = \(\frac n2\) [2×10 + (n - 1)×\(\frac 13\)]

or, 400 = n[\(\frac {20 × 3 + n - 1}{3}\)]

or, 1200 = 60n + n² - n

or, 1200 =n² + 59n

or, n² + 59n - 1200 = 0

or, n² + 75n - 16n - 1200 = 0

or, n(n + 75) - 16(n + 75) = 0

or, (n + 75) (n - 16) = 0

Either: n + 75 = 0 ∴n = -75 (Impossible)

Or: n - 16 = 0 ∴n = 16

∴The number of students = 16 _Ans

Here,

third term (t₃) = 1

no. of terms (n) = 3

first term (a) = ?

common difference (d) = ?

We know,

t_n = a + (n - 1)d

t₃ = a + (3 - 1)d

1 = a + 2d

a = 1 - 2d ....................(1)

Again,

fifth term (t₅) = 7

number of terms (n) = 5

We know,

t_n = a + (n - 1)d

t₅ = a + (5 - 1)d

7 = a + 4d.....................(2)

Putting the value of a in equation (ii)

7 = 1 - 2d + 4d

or, 7 - 1 = 2d

or, d = \(\frac 62\)

∴ d = 3

Putting the value of d in equation (i)

a = 1 - 2× 3 = 1 - 6 = - 5

Again,

a = -5

d = 3

n = 10

S₁₀ = ?

We know,

S_n = \(\frac n2\) [2a + (n - 1) d]

S₁₀ = \(\frac {10}{2}\) [2×(-5) + (10 - 1) 3] = 5[- 10 + 27] = 5×17 = 85 _Ans

Here,

fifth term (t₅) = 17

number of term (n) = 5

tenth term (t₁₀₎ = 42

number of terms (n) = 10

We know,

t_n= a + (n -1)d

t₅ = a + (5 - 1)d

17 = a + 4d ................(1)

t_n= a + (n -1)d

t₁₀= a + (10 -1)d

42 = a + 9d .................(2)

Subtracting eqⁿ(1) from eqⁿ(2)

d = \(\frac {25}{5}\)

∴ d = 5

putting the value of d in eqⁿ (1)

a + 4d = 17

or, a + 4× 5 = 17

or, a = 17 - 20

∴ a = -3

Again,

a = -3

d = 5

n = 12

S₁₂ = ?

S_n = \(\frac n2\) [2a + (n - 1) d]

S₁₂ = \(\frac {12}{2}\) [2(-3) + (12 - 1) 5] = 6[-6 + 55] = 6× 49 = 294 _Ans

Here,

fifth term (t₅) = 19

number of terms (n) = 5

t_n= a + (n - 1)d

t₅ = a + (5 - 1)d

19 = a + 4d ......................(1)

Again,

eighth term (t₈) = 31

number of terms (n) = 8

t_n= a + (n - 1)d

t₈ = a + (8 - 1)d

31 = a + 7d .........................(2)

Subtracting equation (2) from equation (1)

∴ d = \(\frac {-12}{-3}\) = 4

Putting the value of d in eqⁿ(1)

a + 4d = 19

or, a + 4× 4 = 19

or, a + 16 = 19

or, a = 19 - 16

∴ a = 3

last term (l) = 67

first term (a) = 3

common difference (d) = 4

We know,

l = a + (n - 1) d

or, 67 = 3 + (n - 1) 4

or, 67 - 3 = 4n -4

or, 4n - 4 = 64

or, 4n = 64 + 4

or, 4n = 68

or, n = \(\frac {68}{4}\)

∴ n = 17

∴ The 67 number be 17^th term. _Ans

Here,

third term (t₃) = 20

number of terms (n) = 3

t_n=a + (n - 1)d

t₃ = a + (3 - 1)d

20 = a + 2d .......................(1)

9th term (t₉) = 5

number of terms (n) = 9

t_n=a + (n - 1)d

t₉ = a + (9 - 1)d

5 = a + 8d ..........................(2)

subtracting eqⁿ(1) from eqⁿ(2)

or, d = -\(\frac {15}{6}\)

∴ d = -\(\frac 52\)

Putting the value of d in eqⁿ (1)

a + 2d = 20

or, a + 2× \(\frac {-5}{2}\) = 20

or, a - 5 = 20

∴ a = 25

∴ d = \(\frac {-5}{2}\)

19^th term (t₁₉) = ?

number of terms (n) = 19

t_n= a + (n - 1)d

or, t₁₉ = 25 + (19 - 1) × -\(\frac {5}{2}\)

or, t₁₉ = 25 + 18 × -\(\frac {5}{2}\)

or, t₁₉ = 25 - 14

∴ t₁₉ = - 20

∴ 19^th term be - 20. _Ans

Here,

first term (a) = 140

last term (l) = -60

number of arithmetic mean (n) = 4

common difference (d) = \(\frac {b - a}{n + 1}\) = \(\frac {-60 - 140}{4 + 1}\) = \(\frac {-200}{5}\) = -40

Let: m₁, m₂, m₃ and m₄ are the A.M. then,

m₁ = a + d = 140 - 40 = 100

m₂ = a + 2d = 140 - 2 × 40 =140 - 80 = 60

m₃ = a + 3d = 140 - 3 × 40 =140 - 120 = 20

m₄ = a + 4d = 140 - 4 × 40 =140 - 160 = -20

∴ Required 4 A.M. are 100, 60, 20 and - 20 _Ans

Here,

first term (a) = 12

last term (b) = 33

fourth term (m₄) = 24

number of mean (n) = ?

common difference (d) = \(\frac {b - a}{n + 1}\) = \(\frac {33 - 12}{n + 1}\) = \(\frac {21}{n + 1}\)

fourth mean (m₄) = a + 4d

or, 24 = 12 + 4× \(\frac {21}{n + 1}\)

or, 24 - 12 = \(\frac {84}{n + 1}\)

or, 12 = \(\frac {84}{n + 1}\)

or, n + 1 = \(\frac {84}{12}\)

or, n = 7 - 1

∴ n = 6

∴The number of means = 6 _Ans

Here,

first mean (a) = 3

last term (b) = 39

number of terms (n) = ?

\(\frac {third mean}{last mean}\) = \(\frac {m_3}{m_n}\) = \(\frac 37\)

common difference (d) = \(\frac {b - a}{n + 1}\)

d = \(\frac {39 - 3}{n + 1}\)

∴ d = \(\frac {36}{n + 1}\)

From Question,

\(\frac {m_3}{m_n}\) = \(\frac {a + 3d}{a + 4n}\) [\(\because\) m_n = a + nd]

or, \(\frac 37\) =\(\frac {3 + 3 × (\frac {36}{n + 1})}{3 + n × (\frac {36}{n + 1})}\) = \(\cfrac{\frac{3n + 3 + 108}{n + 1}}{\cfrac{3n + 3 + 36n}{n + 1}}\)

or, \(\frac 37\) = \(\frac {3n + 111}{39n + 3}\)

pr, 117n + 9 = 21n + 777

or, 117n - 21n = 777 - 9

or, 96n = 768

or, n = \(\frac {768}{96}\)

∴ n = 8

∴The number of mean = 8 _Ans

Here,

number of terms in A.S (n) = 20

last term (l) = 195

common difference (d) = 5

first term (a) = ?

sum of 20^th term (S₂₀) = ?

We know that,

l = a + (n - 1)d

or, 195 = a + (20 - 1) 5

or, 195 = a + 19× 5

or, a = 195 - 95

∴ a = 100

S_n = \(\frac n2\) [a + l]

S₂₀ = \(\frac {20}{2}\) [100 + 195] = 10× 295 = 2950 _Ans

Here,

first term in A.S (a) = - 24

last term (l) = 72

sum of the series (S_n) = 600

number of terms (n) = ?

common difference (d) = ?

We have,

S_n = \(\frac n2\) (a + l)

or, n = \(\frac {1200}{48}\) = 25

∴ n = 25

Again,

l = a + (n - 1)d

or, 72 = -24 + (25 - 1)d

or, 72 + 24 = 24d

or, d = \(\frac {96}{24}\) = 4

∴ d = 4

∴ number of terms (n) = 25

and common difference (d) = 4 _Ans

Let:

4^th term (t₄) = 4

no. of items (n) = 4

first term (a) = ?

common difference (d) = ?

Using formula,

t_n= a + (n - 1)d

t₄ = a + (4 - 1)d

4 = a + 3d ............(1)

Again,

8^th term (t₈) = 32

no. of terms (n) = 8

first term (a) = ?

common difference (d) = ?

Using formula,

t_n= a + (n - 1)d

t₈ = a + (8 - 1)d

32 = a + 7d ..........(2)

Equation (2) - (1) we get:

or, d = \(\frac {28}{4}\)

∴ d = 7

Putting the value of d in equation (1)

a + 3d = 4

a + 3× 7 = 4

a = 4 - 21

∴ a = -17

Again,

sum of 10^th term (S₁₀) = ?

first term (a) = - 17

number of term (n) = 10

common difference (d) = 7

Using formula,

S_n = \(\frac n2\) [2a + (n - 1) d]

S₁₀ = \(\frac {10}{2}\) [2× -17 + (10 - 1)^7] = 5[-34 + 63] = 5× 29 = 145 _Ans

Here,

Let: m₁, m₂, m₃, m₄ and m₅ are the five A.Ms between 10 and 40.

first term (a) = 10

last term (b) = 5

total number of terms (n) = 5 + 2 = 7

common difference = d

Using formula,

t_n= a + (n - 1)d

40 = 10 + (7 - 1) d

or, 40 - 10 = 6d

or, d = \(\frac {30}{6}\)

∴ d = 5

m₁ = a + d = 10 + 5 = 15

m₂ = a + 2d = 10 + 2× 5 = 10 + 10 = 20

m₃ = a + 3d = 10 + 3 × 5 = 10 + 15 = 25

m₄ = a + 4d = 10 + 4 × 5 = 10 + 20 = 30

m₅ = a + 5d = 10 + 5 × 5 = 10 + 25 = 35

Hence, required A.Ms are 15, 20, 25, 30 and 35. _Ans

Here,

first men (m₁) = 18

fifth mean (m₅) = 46

Let: 18, m₂, m₃, m₄, 46 be 5 A.Ms between a and b then a, 18, m₂, m₃, m₄, 46, b are in Arithmetic sequence.

We know that,

t_n = a + (n - 1) d

m₁= a + (2 - 1)d

18 = a + d ............(1)

m₅= a + (6 - 1)d

46 = a + 5d ..........(2)

Subtracting equation (1) from (2)

or, d = \(\frac {28}{7}\)

∴ d = 7

Substituting the value of d in equation (1)

a + d = 18

a + 7 = 18

a = 18 - 7

∴ a = 11

m₂ = a + 2d = 11 + 2× 7 = 11 + 14 = 25

m₃ = a + 3d = 11 + 3 × 7 = 11 + 21 = 32

m₄ = a + 4d = 11 + 4 × 7 = 11 + 28 = 39

b = a + 6d = 11 + 6 × 7 = 11 + 42 = 53

∴ a = 11, m₂ = 25, m₃ = 32, m₄ = 39 and b = 53 _Ans

Let: the three numbers are in A.P. are a - d, a and a + d.

By question,

Condition 1^st:

a - d + a + a + d =15

or, 3a = 15

or, a = \(\frac {15}{3}\)

∴ a = 5

Condition 2^nd:

(a - d)² + a² + (a - d)² = 83

or, (5 - d)² + 5² + (5 + d)² = 83

or, 25 - 10d + d² + 25 + 25 + 10d + d² = 83

or, 2d²+ 75 = 83

or, 2d² = 83 - 75

or, d² = \(\frac 82\)

or, d² = 4

∴ d = ±2

Taking +ve sign:

a - d = 5 - 2 = 3

a = 5

a + d = 5 + 2 = 7

Taking -ve sign:

a - d = 5 + 2 = 7

a = 5

a + d = 5 - 2 =3

Hence, the A.P. is 3, 5, 7 or, 7, 5, 3. _Ans

Here,

first term (a) = 5

last term (l) = 80

third term (t₃) = 20

common ratio = r

We know that,

t₂ = ar

t₃ = ar²

or, 20 = 5× r²

or, r² = \(\frac {20}{5}\)

or, r² = 4

∴ r = 2

All geometric means between 5 and 80.

m₁ = ar = 5× 2 = 10

m₂ = ar²= 5× 2²= 20

m₃ = ar³= 5× 2³= 40

last term =ar⁴= 5× 2⁴= 80

∴ number of means = 3

and last mean = 40 _Ans

Let: the required number be a and b.

A.M. = \(\frac {a + b}{2}\) = 25

∴ a + b = 50 ..................(1)

G.M. = \(\sqrt {ab}\) = 20

Squaring on bot sides of the above

ab = 400 ....................(2)

We know,

(a - b)² = (a + b)² - 4ab

(a - b)² = (50)² - 4× 400

(a - b)² = 2500 - 1600

(a - b)² = 900

a - b = 30 ..................(3)

Adding equation (1) and (3)

or, a = \(\frac {80}{2}\)

∴ a = 40

Putting the value of a in equation (1)

a + b = 50

or, 40 + b = 50

or, b = 50 - 40

∴ b = 10

Hence, the required numbers are 40 and 10. _Ans

Here,

third term (t₃) = \(\frac 13\)

number of terms (n) = 3

t_n= ar^n-1

t₃= ar^3-1

\(\frac 13\) = ar²...........(1)

sixth term (t6) = \(\frac {1}{81}\)

no. of term (n) = 6

t_n= ar^n-1

t₆= ar^6-1

\(\frac {1}{81}\) = ar⁵ ............(2)

Dividing equation (2) by (1):

\(\frac {ar^5}{ar^2}\) = \(\cfrac{\frac{1}{81}}{\cfrac{1}{3}}\)

or, r³= \(\frac {1}{81}\) × \(\frac 31\)

or, r³= \(\frac {1} {27}\)

or, (r)³ = (\(\frac 13\))³

∴ r = \(\frac 13\)

Putting the value of r in equation (1)

ar² = \(\frac 13\)

or, a× (\(\frac 13\))² = \(\frac 13\)

or, a× \(\frac 19\) = \(\frac 13\)

∴ a = \(\frac 93\) = 3

Again,

a = 3

r = \(\frac 13\)

n = 6

S₆ = ?

We know,

S_n = \(\frac{a(r^n - 1)}{r - 1}\)

=\(\frac{3((\frac 13)^6 - 1)}{\frac 13 - 1}\)

=\(\frac{3(\frac {1}{729} - 1)}{\frac {1-3}{3}}\)

=\(\frac{3(\frac {1 - 729}{729})}{-\frac 23}\)

= 3× \(\frac {-728}{729}\)× \(\frac {-3}{2}\)

= \(\frac {6552}{1458}\)

= 4\(\frac {40}{81}\) _Ans

Here,

second term (t₂) = 4

number of term (n) = 2

t_n= ar^n-1

t₂= ar^2-1

4 = ar ..................(1)

fifth term (t₅) = 32

no. of term (n) = 5

t_n = ar^n-1

t₅ = ar^5-1

32 = ar⁴....................(2)

Eqⁿ (1) is divided by eqⁿ (2)

\(\frac {ar^4}{ar}\) = \(\frac {32}{4}\)

r³ = 8

r = 2

Putting the value of r in eqⁿ (1)

ar = 4

or, a× 2 = 4

or, a = \(\frac 42\)

∴ a = 2

Again,

eight term (t₈) = ?

no. of term (n) = 8

first term (a) = 2

common ratio (r) = 2

t_n= ar^n-1

t₈= 2 × 2^8-1 = 256

∴ The eight term = 256 _Ans

Let: three terms of G.P. are \(\frac ar\), a and ar.

From first condition:

\(\frac ar\) + a + ar = 62 .................... (1)

From second condition:

\(\frac ar\)× a× ar = 1000

or, a³ = 10³

i.e. a = 10

Putting the value of a in eqⁿ (1),

\(\frac ar\) + a + ar = 62

or, \(\frac {10}{r}\) + 10 + 10r = 62

or, \(\frac {10 + 10r}{r}\) = 62 - 10

or, 10 + 10r² = 52r

or, 10r² - 52r + 10 = 0

or, 2(5r² -26r + 5) = 0

or, 5r² - 26r +5 = 0

or, 5r(r - 5) -1(r - 5) = 0

or, (r - 5) (5r - 1) = 0

Either: r - 5 = 0 ∴r = 5

Or, 5r - 1 = 0 ∴r = \(\frac 15\)

If r = 5, a = 10 then

\(\frac ar\) = \(\frac {10}{5}\) = 2

ar = 10× 5 = 50

If r = \(\frac 15\) and a = 10 then

\(\frac ar\) =\(\cfrac{10}{\cfrac{1}{5}}\) = 50

ar = 10× \(\frac 15\) = 2

∴ Required numbers are 2, 10 and 50 or 50, 10 and 2. _Ans

Let: the three number in A.P. be a - d, a, a + d

From first condition:

a - d + a + a + d = 15

or, 3a = 15

or, a = \(\frac {15}{3}\)

∴ a = 5

From second condition:

If 1, 3, 9 be added then a - d + 1, a + 3, a + d + 9 are in G.P.

5 - d + 1, 5 + 3, 5 + d + 9 are in G.P.

6 - d, 8, 14 + d are in G.P.

or, \(\frac {8}{6 - d}\) = \(\frac {14 + d}{8}\)

or, 64 = (6 - d) (14 + d)

or, 84 + 6d - 14d - d² = 64

or, -d² - 8d + 84 - 64 = 0

or, -(d² + 8d - 20) = 0

or, d² - 2d + 10d - 20 = 0

or, d(d - 2) + 10(d - 2) = 0

or, (d - 2) (d + 10) = 0

Either: d - 2 = 0∴ d = 2

Or: d + 10 = 0 ∴d = -10

When, a = 5 and d = 2 then

a - d, a, a + d

5 - 2, 5, 5 + 2

3, 5, 7

When, a = 5 and d = - 10

a - d, a, a + d

5 - (-10), 5, 5 + (-10)

5 + 10, 5, 5 - 10

15, 5, -5

∴ The three numbers are 3, 5, 7 or 15, 5, -5 _Ans

Let: the two number are a and b.

From question:

\(\frac {A.M}{G.M}\) = \(\frac 53\)

We know that,

A.M. = \(\frac {a + b}{2}\)

G.M. = \(\sqrt {ab}\)

\(\cfrac{\frac {a + b}{2}}{{\sqrt {ab}}}\) = \(\frac 53\)

or, \(\frac {3(a + b)}{2}\) = 5\(\sqrt {ab}\)

or, 3(a + b) = 10\(\sqrt {ab}\)

Squaring on both sides,

[3(a + b)]² = (10\(\sqrt {ab}\))²

or, 9(a² + 2ab + b²) = 100ab

or, 9a² + 18ab + 9b²- 100ab = 0

or, 9a²- 82ab + 9b² = 0

or, 9a² - 81ab - ab + 9b² = 0

or, 9a(a - 9b) -b(a - 9b) = 0

or, (a - 9b) (9a - b) = 0

Either: a - 9b = 0

a = 9b

\(\frac ab\) = \(\frac 91\)

Or: 9a - b = 0

9a = b

\(\frac ab\) = \(\frac 19\) (impossible)

∴ The required ratio is 9 : 1. _Ans

Let: the three terms of the G.P. be \(\frac ar\), a and ar.

From first condition:

\(\frac ar\) + a + ar = 28 ..........................(1)

From second condition:

\(\frac ar\) × a × ar = 512

or, a³ = 8³

∴ a = 8

Putting the value of a in eqⁿ (1)

\(\frac ar\) + a + ar = 28

or, \(\frac 8r\) + 8 + 8r = 28

or, \(\frac {8 + 8r^2}{r}\) = 28 - 8

or, 8 + 8r² = 20r

or, 8r² - 20r + 8 = 0

or, 4(2r² - 5r + 2) = 0

or, 2r² - 5r + 2 = 0

or, 2r² - 4r - r + 2 = 0

or, 2r(r - 2) - 1(r - 2) = 0

or, (r - 2)(2r - 1) = 0

Either: r - 2 = 0 ∴r = 2

Or: 2r - 1 = 0 ∴r = \(\frac 12\)

When, r = \(\frac 12\) and a = 8 then:

\(\frac ar\), a, ar

=\(\cfrac{8}{\cfrac{1}{2}}\), 8, 8× \(\frac 12\)

= 16, 8, 4

When, r = 2 and a = 8 then:

\(\frac ar\), a, ar

= \(\frac 82\), 8, 8× 2

= 4, 8, 16

∴ The no. are 4, 8, 16 or 16, 8, 4. _Ans

Let: the four consecutive terms are a, ar, ar² and ar³.

From first condition:

a + ar + ar² + ar³ = 30

or, a(1 + r) + ar²(1 + r) = 30

or, (1 + r) (a + ar²) = 30

or, a(1 + r) (1 + r²) = 30 .........................(1)

From second condition:

\(\frac {a + ar^3}{2}\) = 9

or, a(1 + r³) = 18

or, a(1 + r) (1 - r + r²) = 18

or, a(1 + r) = \(\frac {18}{1 - r + r^2}\) ..........................(2)

Putting the value of a(1 + r) in eqⁿ (1)

\(\frac {18}{1 - r + r^2}\) (1 + r²) = 30

or, 18 + 18r² = 30 - 30r + 30r²

or, 30r² - 30r + 30 - 18 - 18r² = 0

or, 12r² - 30r + 12 = 0

or, 6(2r² - 5r + 2) = 0

or, 2r² - 5r + 2 = 0

or, 2r(r- 2) -1(r - 2) = 0

or, (r - 2)(2r - 1) = 0

Either: r - 2 = 0 ∴r = 2

Or: 2r - 1 = 0 ∴r = \(\frac 12\)

∴r = 2 or \(\frac 12\) _Ans

Here,

Given series is: 3 + 6 + 12 + .......... + 768

common ratio (r) = \(\frac 63\) = \(\frac {12}{6}\) = 2. So it is a geometric series.

First term (a) = 3

common ratio (r) = 2

last term (l) = 768

(a) no. of terms of the sequence (n) = ?

We know that,

l = ar^n-1

or, 768 = 3× 2^n-1

or, 2^n-1 = \(\frac {768}{3}\)

or, 2^n-1 = 256

or, 2^n-1 = 2⁸

or, n - 1 = 8

∴ n = 8 + 1 = 9

no. of terms (n) = 9

(b) The sum of the series (S₉) = ?

We know,

S_n = \(\frac {lr - a}{r - 1}\)

S₉ = \(\frac {768 × 2 - 3}{2 - 1}\) = \(\frac {1536 - 3}{1}\) = 1533

∴ The sum of the series = 1533 _Ans

Here,

4^th term (t₄) = \(\frac 12\)

number of terms (n) = 4

We know that,

t_n = ar^n-1

t₄ = ar^4-1

\(\frac 12\) = ar³.............(1)

9^th term (t₉) = \(\frac {16}{243}\)

number of terms (n) = 5

We know,

t_n = ar^n-1

t₉ = ar^9-1

\(\frac {16}{243}\) = ar⁸..................(2)

Dividing equation (2) by (1): we get,

\(\cfrac{\frac{16}{243}}{\cfrac{1}{2}}\) = \(\frac {ar^8}{ar^3}\)

or, \(\frac {16× 2}{243}\) = r⁴

or, r⁵ = \(\frac {2^5}{3^5}\)

or, (r)⁵ = (\(\frac 23\))⁵

∴ r = \(\frac 23\)

Substituting the value of r in equation (1)

ar³ = \(\frac 12\)

or, a(\(\frac 23\))³ = \(\frac 12\)

or, a× \(\frac {8}{27}\) = \(\frac 12\)

∴ a = \(\frac {27}{16}\)

Again,

fifth term (t₅) = ar^5-1

= \(\frac {27}{16}\)× (\(\frac 23\))⁴

= \(\frac {27}{16}\)× \(\frac {16}{27}\)× \(\frac 13\)

= \(\frac 13\) _Ans

Here,

third term (t₃) = \(\frac 14\)

number of terms (n) = 3

We know that,

t_n = ar^n-1

t₃ = ar^3-1

\(\frac 13\) = ar² ......................(1)

fourth term (t₄) = \(\frac 18\)

number of terms (n) = 4

t_n = ar^n-1

t₄ = ar^4-1

\(\frac 18\) = ar³ ............................(2)

Dividing (2) by (1),

\(\frac {ar^3}{ar^2}\) =\(\cfrac{\frac{1}{8}}{\cfrac{1}{4}}\)

or, r = \(\frac 18\) ×\(\frac 41\) = \(\frac 12\)

Substituting the value of r in equation (1)

ar² = \(\frac 14\)

a(\(\frac 12\))² = \(\frac 14\)

a = \(\frac 14\) ×\(\frac 41\)

a = 1

Again,

S_n = \(\frac {a( r^n - 1)}{ r - 1}\)

S₈ = \(\frac {1[(\frac {1}{2})^8 - 1)]}{\frac 12 - 1}\)

S₈ =\(\cfrac{\frac{1}{256} - 1}{\cfrac{1}{2} - 1}\)

S₈=\(\cfrac{\frac{1 - 256}{256}}{\cfrac{1 - 2}{2}}\)

S₈ = \(\frac {-255}{256}\)× \(\frac {2}{-1}\)

S₈ = \(\frac {255}{128}\)

∴ The sum of eight terms = 1\(\frac {127}{128}\) _Ans

Here,

Let: the three terms be \(\frac ar\), a and ar.

From first condition,

\(\frac ar\) + a + ar = 21 .................(1)

From second condition,

\(\frac ar\) ×a ×ar = 64 ............................(2)

or, a³ = 4³

∴ a = 4

Substituting the value of a in equation (1)

\(\frac ar\) + a + ar = 21

or, \(\frac 4r\) + 4 + 4r = 21

or, \(\frac 4r\) + 4r = 21 - 4

or, \(\frac {4 + 4r^2}{r}\) = 17

or, 4r² + 4 = 17r

or, 4r² - 17r + 4 = 0

or, 4r² - 16r - r + 4 = 0

or, 4r(r - 4) - 1(r - 4) = 0

or, (r - 4) (4r - 1) = 0

Either: r - 4 = 0 ∴r = 4

Or: 4r - 1 = 0 ∴ r = \(\frac 14\)

If a = 4 and r = 4

first term = \(\frac ar\) = \(\frac 44\) = 1

second term = a = 4

third term = ar = 4× 4 = 16

If a = 4 and r = \(\frac 14\)

first term =\(\cfrac{4}{\cfrac{1}{4}}\) = 16

second term = a = 4

third term = 4 ×\(\frac 14\) = 1

Hence, the number are 1, 4 and 16 or 16, 4 and 1. _Ans

Here,

first term (a) = \(\frac {1}{16}\)

last term (b) = 64

number of means (n) = 4

We know that,

r = (\(\frac ba\))^{\(\frac{1}{n+1}\)}

r = (\(\cfrac{64}{\cfrac{1}{16}}\))^{\(\frac{1}{n+1}\)}= (64 × 16)^{\(\frac 15\)} = (2⁶× 2⁴)^{\(\frac 15\)}=(2)^{10×\(\frac 15\)} = 2²= 4

If G₁, G₂, G₃ and G₄are the 4 geometric means.

G₁ = ar = \(\frac {1}{16}\)× 4 = \(\frac 14\)

G₂ = ar² = \(\frac {1}{16}\)× 4× 4 = 1

G₃ = ar³ = \(\frac {1}{16}\)× 4× 4 × 4 =4

G₄ = ar⁴ = \(\frac {1}{16}\)× 4× 4 × 4 × 4= 16

∴ The required means are \(\frac14\), 1, 4 and 16. _Ans

Let: a, b and c are in A.P.

b - a = c - b

or, b + b = c + a

or, 2b = c + a

∴ b = \(\frac {c + a}{2}\) .......................(1)

Again,

x, y and z are in G.P.

\(\frac yx\) = \(\frac zy\)

or, y² = xz ...........................(2)

L.H.S.

=x^b-c× y^c-a× z^a-b

= x^a-b× y^c-a× z^a-b[\(\because\) a - b = b - c]

= (xz)^a-b× y^c-a

= (y²)^a-b× y^c-a

= y^2a-2b+c-a

= y^a+c-2b

= y^2b-2b

= y⁰

= 1

= R.H.S _proved