Sequences and Series

11 )

If two arithmetic means A₁, A₂, two geometric means G₁, G₂ and two harmonic means H₁, H₂ are inserted between two numbers P and Q then ____

[(G1G2) / (H1H2)] = [(A1 + A2) / (H1 + H2)]

[(G1 + G2) / (H1 + H2)] = [(A1A2) / (H1H2)]

[(G1G2) / (H1H2)] = [(A1 – A2) / (H1 – H2)]

(A1 + A2) (H1 + H2) = G1G2H1H2

View Solution

Answer :

Here P, A₁, A₂ Q are in A. P. ⇒ A₁ + A₂ = P + Q

P, G, G₂, Q are in G. P. ⇒ G₁G₂ = PQ

also P, H₁, H₂ Q are in H. P. ⇒ (1 / H₁) – (1 / P) = (1 / Q) – (1 / H₂)

∴ (1 / H₁) + (1 / H₂) = (1 / P) + (1 / Q)

∴ [(H₁ + H₂) / (H₁H₂)] = [(P + Q) / (PQ)]

= [(A₁ + A₂) / (G₁G₂)]

∴ [(G₁G₂) / (H₁H₂)] = [(A₁ + A₂) / (H₁ + H₂)]

12 )

∑(∑2^1–n) = _____

Select correct option from following options

(A) 2n – 2 + n^1–n

(B) 2n – 2 + n^n–1

(D) 2n – 2 + nⁿ⁺¹

View Solution

Answer :

∑(∑2^1–n) = ∑[2{1 – (1 / 2ⁿ)}]

= 2[∑1 – ∑(1 / 2ⁿ)]

= 2[n – {(1/2) + (1/4) + (1/8) + --- (1 / 2ⁿ)}]

= 2n – 2[(1/2) + (1/4) + (1/8) + --- (1 / 2ⁿ)]

= 2n – [1 + (1/2) + (1/4) + --- (1 / 2^n–1)]

= 2n – [{1 – (1 / 2ⁿ)} / {1 – (1/2)}]

= 2n – 2[1 – (1 / 2ⁿ)] = 2n – 2 + 2^1–n

13 )

2 + 6 + 12 + 20 + 30 + … 100 terms = __________

[(1020300) / 3]

[(1030200) / 3]

[(1003200) / 3]

[(1023200) / 3]

View Solution

Answer :

2 + 6 + 12 + 30 + ---- 100 terms

= (1² + 1) + (2² + 2) + (3² + 3) + ----- 100 terms

= [∑n² + ∑n]_n=100

= [{n(n + 1)(2n + 1)} / 6] + [{n(n + 1)} / 2]

= [(100 × 101 × 201) / 6] + [(100 × 101) / 2]

= [(1030200) / 3]

14 )

If any terms of an A. P. is non – zero and d ≠ 0 then ^n–1∑_r=1 [1 / (a_r a_r+1)] = _________

Select correct option from following options

(A) [n / (a₁ a_n)] (B) [(n – 1) / (a₁ a_n)]

View Solution

Answer :

^n–1∑_r=1 [1 / (a_r ∙ a_{r + 1})] = (1/d) ^n–1∑_r=1 [d / (a_r ∙ a_{r + 1})]

= (1/d) ^n–1∑_r=1 [(a_{r + 1} – a_r) / (a_r ∙ a_{r + 1})]

∴ ^n–1∑_r=1 [1 / (a_r ∙ a_{r + 1})] = ^n–1∑_r=1 [(1 / a_r) – {1 / (a_{r + 1})}] ∙ (1/d)

= (1/d) [(1 / a₁) – (1 / a₂) + (1 / a₂) – (1 / a₃) + --- (1 / a_n–1) – (1 / a_n)]

= (1/d) [(1 / a₁) – (1 / a_n)]

= (1/d) [(a_n – a₁) – (a₁ ∙ a_n)]

= (1/d) [{(n – 1)d} – (a₁ a_n)]

= [(n – 1) / (a₁ ∙ a_n)]

15 )

If S₁, S₂, S₃, … S_n are the sums of infinite G. P. s. whose first terms are 1, 2, 3, …, n and whose common ratios are (1/2), (1/3), (1/4),… [1 / (n + 1)] respectively, then ⁿ∑_i=1 S_i = ______

[{n(n + 3)} / 2]

[{n(n + 4)} / 2]

[{n(n – 3)} / 2]

[{n(n + 1)} / 2]

View Solution

Answer :

ⁿ∑_i=1 S_i = ⁿ∑_i=1 = [a / (1 – r)] = [1 / {1 – (1/2)}] + [2 / {1 – (1/3)}] + [3 / {1 – (1/4)}] + ----- [n / {1 – {1 / (n + 1)}}] = 2 + 3 + 4 + ---- + (n + 1)

= (n/2)[4 + (n – 1)1]

= [{n(n + 3)} / 2]

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