CRE 1 - Arithmetic Progression | Algebra - Progressions

Answer: Option C

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Explanation :

The given series 2√3 + √3 + 0 + … is an A.P. where

First term (a) = 2√3 and 

Common difference (d) = - √3

nth term of an AP (Tn) = a + (n – 1)d

∴ 10^th term of the series = 2√3 + (10 - 1)(-√3) = -7√3

Hence, option (c)

Answer: Option D

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Explanation :

Given that 11^th term is zero

⇒ a + (11 – 1) d = 0

⇒ a + 10d = 0.

⇒ a = -10d

Now ratio of 5^th and 7^th terms =

$\frac{a+4d}{a+6d}=\frac{-10d+4d}{-10d+6d}=\frac{-6d}{-4d}=\frac{3}{2}$

Hence, option (d).
 

Answer: 134

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Explanation :

Give series is an AP with a = 2 and d = 3.

n^th term of the series (A.P.) ⇒ T_{​​​​​​​n} = a + (n - 1)d

So, 401 = 2 + (n – 1) × 3

⇒ n = 134

Hence, 134.

Answer: Option B

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Explanation :

Let the first term of the series = a and common difference = d.

Given that, T_p = a + (p – 1)d = q            …(1)

And T_q = a + (q – 1)d = p            …(2)

(2) - (1), we get

(p - q)d = q - p

⇒ d = $\frac{q-p}{(p-q) }$ = -1

Putting the value of d in equation (1), then a = p + q – 1 

Now r^th term is given by A.P.

T_r = a + (r - 1)d = (p + q - 1) + (r - 1)(-1) = p + q – r.

Alternately,
Put p = 1, q = 2 and r = 3.

∴ T₁ = 2 and T2 = 1.

⇒ a = 2 and d = -1.

∴ T₃ = 2 + (3 - 1) × -1 = 0

Use options now. By substituting p = 1, q = 2 and r = 3 and see which options gives us T₃ = 0.

Option (b) gives us p + q – r = 1 + 2 – 3 = 0.

Hence, option (b).

Answer: Option C

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Explanation :

The given numbers will form an Arithmetic Series i.e., 12 + 15 + 18 + ……. 198 and 

the number of terms is = $\frac{{\mathrm{T}}_{\mathrm{n}}-{\mathrm{T}}_1}{\mathrm{d}}$ + 1 = $\frac{198-12}{3}$ + 1 = 63

Now sum = average × number of terms = $\frac{12+198}{2}$ × 63 = 6615

Hence, option (c).

Answer: Option A

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Explanation :

Let the sides of the triangle be a –d, a, a + d, then hypotenuse being the greatest side i.e. a + d.

So (a + d)² = a² + (a – d)²

⇒ a² + d² + 2ad = 2a² - 2ad + d²

⇒ a = 4d.

Therefore ratio of the sides = a – d : a : a + d = (4d – d) : 4d : (4d + d) = 3 : 4 : 5.

Hence, option (a).

Answer: Option C

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Explanation :

The given number are 13, 17 …….. 97. 

This is an A.P. with first term 13 and common difference 4. 

Let the number of terms be n. 

Then 97 = 13 + (n – 1)4 

⇒ 4n = 88 

⇒ n = 22

∴ Sum of the numbers = n/2 × [First term + Last term] = 22/2 × [13 + 97] = 11 × 110 = 1210.

Hence, option (c).

Answer: Option A

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Explanation :

Maximum sum will occur when last term is 0 or 2 (since after that terms will be -ve which will start reducing the sum)

∴ Last term = 0 = 52 + (n – 1)(-2) 

⇒ n = 27

∴ S_max = 27/2 × [52 + 0] = 702.

Hence, option (b).

Answer: 17

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Explanation :

By observation we can see the common terms for the two sequences are 5, 11, 17, …

i.e., the common terms form an Arithmetic Progression whose first term is 5 and the common difference is LCM of the common differences of original series i.e., 6.

Let the number of terms in this common series be n.

Last term of this series = 5 + (n - 1) × 6

The highest term of this common series should be less than or equal to the least of last terms of the two series.

i.e., 5 + (n - 1) × 6 ≤ 101

⇒ 6(n - 1) ≤ 96

⇒ n – 1 ≤ 16

⇒ n ≤ 17.

Hence, number of terms in the common series is 17.

Hence, 17.

Answer: Option B

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Explanation :

Given, S₂₀ = 670 = 20/2 × (2a + (20 - 1) × d)

⇒ 2a + 19d = 67   …(1)

Also, S₃₀ = 670 + 785 = 30/2 × (2a + (30 - 1) × d)

⇒ 2a + 29d = 97   …(2)

Solving (1) and (2) we get,

d = 3 and a = 5.

∴ T₁₀ = 5 + (10 - 1) × 3 = 32.

Hence, option (b).

Answer: 325

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Explanation :

Let S = (-50) + (-45) + (-40) + ... + 60 + 65  + 70 + 75

= (-50) + (-45) + (-40) + ...  + 40 + 45 + 50 + 55 + 60 + 65  + 70 + 75

= 55 + 60 + 65 + 70 + 75

= 325

Hence, 325.

Answer: Option D

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Explanation :

Let the first term, common difference, 24^th term and 75^th term be a₁, d, a₂₅, and a₇₅ respectively.

∴ a₂₅ = a₁ + 24d = 25 and a₇₅ = a₁ + 74d = 75

∴ d = 1, a₁ = 1 

∴ T_{99​​​​​​​} = 99

∴ Average of 99 terms = (a + T₉₉)/2 = (1 + 99)/2 = 50

Hence, option (d).

Answer: Option B

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Explanation :

Let the first term be a and common difference be d.

⇒ T_x = a + (x - 1)d, and
⇒ T_y = a + (y - 1)d

Given, xT_x = yT_y

⇒ x(a + (x - 1)d) = y(a + (y-1)d)
⇒ ax + x²d - xd = ay + y²d - yd
⇒ a(x - y)  + d(x² - y²) - d(x - y) = 0
⇒ (x - y)(a + (x + y)d - d) = 0
⇒ a + (x + y - 1)d = 0 = T_x+y

∴ (x + y)^th term is zero.

Hence, option (b).

Answer: Option A

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Explanation :

Since p, q and r are in AP, let p = a, q = a + d and r = a + 2d 

Now, $\frac{1}{\mathrm{qr}}$ = $\frac{1}{(\mathrm{a}+\mathrm{d})(\mathrm{a}+2\mathrm{d})}$, 

$\frac{1}{\mathrm{pr}}$ = $\frac{1}{\left(\mathrm{a}\right)(\mathrm{a}+2\mathrm{d})}$ and 

$\frac{1}{\mathrm{pq}}$ = $\frac{1}{\left(\mathrm{a}\right)(\mathrm{a}+\mathrm{d})}$

Here, $\frac{1}{\mathrm{qr}}$ + $\frac{1}{\mathrm{pq}}$ = $\frac{1}{(\mathrm{a}+\mathrm{d})(\mathrm{a}+2\mathrm{d})}$ + $\frac{1}{\left(\mathrm{a}\right)(\mathrm{a}+\mathrm{d})}$ = $\frac{\left(a\right)+(a+2d)}{(\mathrm{a}+\mathrm{d})(a+d)(\mathrm{a}+2\mathrm{d})}$ = $\frac{2}{\left(\mathrm{a}\right)(\mathrm{a}+2\mathrm{d})}$ = $\frac{2}{\mathrm{pr}}$

∴ $\frac{1}{\mathrm{qr}}$, $\frac{1}{\mathrm{pr}}$ and $\frac{1}{\mathrm{pq}}$ are in AP.

Hence, option (a).