Sequences and Series

6 )

If the A. M. of two numbers a and b is equal to √(10) times their G. M. then [(a – b) / (a + b)] = __________

[âˆš(10) / 3]

3âˆš(10)

[9 / (10)]

[3 / âˆš(10)]

View Solution

Answer :

∵ AM = √10 GM

∴ [(a + b) / 2] = √(10) √(ab)

∴ (a + b)² = 40 ab

∴ a² + b² = 38 ab ⇒ [(a² + b²) / (2ab)] = 19

∴ [(a² + b² – 2ab) / (a² + b² + 2ab)] = [(19 – 1) / (19 + 1)]

∴ [(a – b) / (a + b)]² = (18 / 20) = (9 / 10)

∴ [(a – b) / (a + b)] = [3 / √(10)]

7 )

If the harmonic mean and geometric mean of two numbers a and b are 4 and 3√2 respectively then the interval [a, b] =___

[3, 6]

[2, 7]

[4, 5]

[1, 8]

View Solution

Answer :

H. M. = 4 and G. M. = 3√2

∴ [(2ab) / (a + b)] = 4 and ab = (G. M.)² = 18

∴ 2(18) = 4(a + b)

∴ a + b = 9

∴ a = 3, b = 6

∴ [a, b] = [3, 6]

8 )

A. M of the three numbers which are in G. P. is [(14) / 3]. If adding 1 in first and second number and subtracting 1 from the third number, resulting numbers are in A. P. then the sum of the squares of original three numbers is____

View Solution

Answer :

Let three numbers in G. P. be

a, ar, ar².

∵ A. M. = (14 / 3)

∴ [(a + ar + ar²) / 3] = (14 / 3)

∴ a + ar + ar² = 14 (1)

also a + 1, ar + 1, ar² – 1 are in A. P.

∴ a + 1 + ar² – 1 = 2(ar + 1)

∴ a + ar² = 2ar + 2

∴ ar² – 2ar + a = 2 (2)

(1) & (2) gives

[(1 + r + r²) / (1 – 2r + r²)] = 7

∴ 1 + r + r² = 7 – 14r + 7r²

∴ 6r² – 15r + 6 = 0

∴ 2r² – 5r + 2 = 0

∴ r = 2 or r = (1/2)

when r = 2, then a = 2 ---- from (2)

∴ numbers are a, ar, ar² i.e. 2, 4, 8.

∴ a² + (ar)² + (ar²)² = 2² + 4² + 8²

= 84

9 )

If the H. M. of a and c is b, G. M. of b and d is c and A. M. of c and e is d, then the G. M. of a and e is__________

View Solution

Answer :

given: b = [(2ac) / (a + c)],

c² = bd and [(c + e) / 2] = d

hence c² = [(2ac) / (a + c)] [(c + e) / 2]

∴ c = [(2a) / (a + c)] [(c + e) / 2] = [(ac + ae) / (a + c)]

∴ ac + c² = ac + ae

∴ c² = ae

∴ Geometric mean of a and e is c.

10 )

If a, b, c are in A. P. and geometric means of ac and ab, ab and bc, ba nad cb are d, e, f respectively then d², e², f² are in ____

A. P.

G. P.

H. P.

A. G P.

View Solution

Answer :

given: ac ∙ ab = d² ⇒ a²bc = d²

bc ∙ ba = e² ⇒ ab²c = e²

ca ∙ cb = f² ⇒ abc² = f²

also a, b, c are in A. P.

∴ a(abc) , b(abc), c(abc) are in A. P.

∴ a²bc, ab²c, abc² are in A. P.

∴ d², e², f² are in A. P.

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