PE 4 - Progression | Algebra - Progressions

Answer: Option B

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Explanation :

Sum of first n natrual numbers = n(n + 1)/2

The least value of n for which the sum is greater 7370 is 121.

∴ The sum of all the numbers should have been (121 × 122)/2 = 7381

∴ The sum of two consecutive numbers that she missed = 7381 - 7370 = 11

⇒ The two consecutive numbers are 5 and 6.

Hence, option (b).

Answer: Option A

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Explanation :

The given sequence is: $\frac{3}{12}$ + $\frac{3}{16}$ + $\frac{3}{20}$+ ...

The given sequence is a Harmonic Progression since the reciprocals are in Arithmetic Progression whose first term is 4 and common difference is 4/3.

Let us consider the AP: $\frac{12}{3}$ + $\frac{20}{3}$ + $\frac{20}{3}$+ ...
T₄ = 20/3 + 4/3 = 24/3 = 8
T₁₃ = $\frac{12}{3}$ + (13 - 1)$\frac{4}{3}$ = $\frac{60}{3}$ = 20

Now, for the HP, 
T₄ = 1/8
T₁₃ = 1/20

∴ Required ratio = T₄ : T₁₃ = 1/8 : 1/20 = 20 : 8 = 5 : 2

Hence, option (a).

Answer: 3

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Explanation :

Let the integers be a, ar and ar², then a + ar + ar² = 111
∴ a(1 + r + r²) = 111
We note that (1 + r + r²) > 0 , hence a > 0
∴ a(1 + r + r²) = (1 × 111), or (3 × 37), or (37 × 3) or (111 × 1)

So, we have the following four cases:

Case 1: When a = 1
1 + r + r² = 111
∴ r² + r − 110 = 0
∴ (r + 11)(r − 10) = 0
∴ r = 10 or −11
So, we get two sets of solutions (1, 10, 100) and (1, −11, 121).

Case 2: When a = 3
1 + r + r² = 37
∴ r² + r − 36 = 0
This has no integral solution.

Case 3: When a = 37
1 + r + r² = 3
∴ r² + r − 2 = 0
∴ (r − 1)(r + 2) = 0
∴ r = 1 or −2
So, we get two sets of solutions (37, 37, 37) and (37, −74, 148). The first set is to be rejected as the integers are not distinct.

Case 4: When a = 111
1 + r + r² = 1
∴ r² + r  = 0
∴ r(r + 2) = 0
∴ r = 0 or −1
So, we get two set of solutions (111, 0, 0) and (111, −111, 111). Both are to be rejected as the integers are not distinct.

∴ From the above four cases, we have 3 distinct valid sets.

Hence, 3.

Answer: Option A

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Explanation :

Let S = 2 + 22 + 222 + ... (upto 10 terms)

⇒ S = 2 × [1 + 11 + 111 + ...]

Now, multiply and divide RHS by 9.

⇒ S = $\frac{2}{9}$ × [9 + 99 + 999 + ...]

⇒ S = $\frac{2}{9}$ × [(101 - 1) + (102 - 1) + (103 - 1) + ... + (1010 - 1)]

⇒ S = $\frac{2}{9}$ × [(101 + 102 + 103 + ... + 1010) - 10]

⇒ S = $\frac{2}{9}$ × $\left[10\left(\frac{{10}^{10}-1}{10-1}\right)-10\right]$

⇒ S = $\frac{2}{9}$ × $\left[10\left(\frac{{10}^{10}-1}{9}\right)-10\right]$

⇒ S = $\frac{2}{9}$ × $\left[\frac{{10}^{11}-100}{9}\right]$

Hence, option (a).

Answer: 365

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Explanation :

The given sequence is a combination of two APs.

The first AP is (2, 5, 8, …) and the second AP is (3, 7, 11, … )

∴ Sum = 10/2 × (2 × 2 + 9 × 3) + 10/2 × (2 × 3 + 9 × 4) = 365

Hence, 365.

Answer: Option E

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Explanation :

Let the first term and common difference of first AP be a₁ and d₁, while that of second AP be a₂ and d₂.

⇒ $\frac{{\mathrm{a}}_1+4{\mathrm{d}}_1}{{\mathrm{a}}_2+4{\mathrm{d}}_2}$ = $\frac{{\mathrm{a}}_1+9{\mathrm{d}}_1}{{\mathrm{a}}_2+9{\mathrm{d}}_2}$

⇒ a₁a₂ + 9a₁d₂ + 4a₂d₁ + 36d₁d₂ = a₁a₂ + 4a₁d₂ + 9a₂d₁ + 36d₁d₂

⇒ 5a₁d₂ = 5a₂d₁

⇒ d₁ : d₂ = a₂ : a₂

Hence, option (e).

Answer: Option C

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Explanation :

T₁ = x and T₂ = y,

⇒ common difference (d) = T₂ - T₁ = y - x, and

Number of terms = $\frac{{\mathrm{T}}_{\mathrm{n}}-{\mathrm{T}}_1}{\mathrm{d}}$ + 1 = $\frac{2\mathrm{x}-\mathrm{x}}{\mathrm{y}-\mathrm{x}}$ + 1 = $\frac{\mathrm{x}}{\mathrm{y}-\mathrm{x}}$ + 1 = $\frac{\mathrm{y}}{\mathrm{y}-\mathrm{x}}$

∴ Sum of all the terms = $\frac{\mathrm{n}}{2}\left[{\mathrm{T}}_1+{\mathrm{T}}_{\mathrm{n}}\right]$ = $\frac{\mathrm{y}}{2(\mathrm{y}-\mathrm{x})}\left[\mathrm{x}+2\mathrm{x}\right]$ = $\frac{3\mathrm{xy}}{2(\mathrm{y}-\mathrm{x})}$

Hence, option (c).

Answer: Option C

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Explanation :

Sum of first a terms is X
Sum of first 2a terms is Y

∴ Out of first 2a terms, sum of last a terms = Y - X

Now, out of first 3a terms, 
sum of first a terms is X
sum of middle a terms is Y - X
let sum of last a terms be S

⇒ sum of last a terms - sum of middle a terms = sum of middle a terms - sum of first a terms
⇒ S - (Y - X) = (Y - X) - X
⇒ S = 2Y - 3X

∴ Sum of all 3 a terms = X + (Y - X) + (2Y - 3X) = 3Y - 3X = 4X - 3X = X [Y = 4X/3]

Hence, option (c).

Answer: Option A

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Explanation :

Let S = 1² - 2² + 3² - 4² + ... + 49² - 50²

Adding and subtracting (2² + 4² + 6² + ... + 50²)

∴ S = (1² + 2² + ... + 50²​​​​​​​) - 2(2² + 4² + 6² + ... + 50²)

⇒ S = (1² + 2² + ... + 50²​​​​​​​​​​​​​​) - 2 × 2² × (1² + 2² + 3² + ... + 25²)

⇒ S = $\frac{50\times 51\times 101}{6}$ - 8 × $\frac{25\times 26\times 51}{6}$

⇒ S = 42925 - 44200 = -1275

Hence, option (a).

Answer: Option B

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Explanation :

Consider 5 numbers p, q, r, s & t such that p, q and r are in A.P., q, r and s are in G.P. and r, s and t are in H.P. p, r and t will form a/an

Since r, s and t are in HP. let us take r = 1/1, s = 1/2 and t = 1/3
[1, 2 and 3 are in AP, hence 1/1, 1/2 and 1/3 are in HP.]

Since, q, r and s are in GP, s = 1/2 and r = 1, hence q = 2
∴ q = 2, r = 1 and s = 1/2

Since, p, q and r are in AP, r = 1 and q = 2, hence p = 3
∴ p = 3, q = 2 and r = 1

∴ p = 3, q = 2, r = 1, s = 1/2 and t = 1/3

Now, p = 3, r = 1 and t = 1/3
⇒ p, r and t form a GP.

Hence, option (b).