Test: Sets and Sequences - ACT MCQ

Concept:
Sum of n terms of an A.P. , S_n = n/2(2a+(n−1)d)
Where a = First term, n = Number of terms and, d = Common Difference.

Calculation:
We need to Find the time taken to count all notes.
Since, a₁ = a₂ = .... = a₁₀ = 150.
Thus, The number of notes counted in first 10 minutes = 150 × 10 = 1500
Now, We have left notes = total number of notes - Number of notes counted in first 10 minutes.
i.e. 4500 - 1500 = 3000
Suppose the person counts the remaining 3000 currency notes in n minutes then,
3000 = Sum of n terms of an AP. with first term 148 and common difference -2 
3000 = n/2(2(148) + (n − 1)( − 2))
⇒ 3000 = n/2(296−2n + 2)
⇒ 3000 = n/2(298−2n)
⇒ 3000 = n(149 - n)
⇒ 3000 = 149n - n²
⇒ n² - 149n - 3000 = 0
⇒ (n - 125)(n - 24) = 0 (By Separating terms)
⇒ n - 125 = 0 and n - 24 = 0
⇒ n = 125, 24
Since, 125 is not possible.
Thus, n = 24
Therefore,
Total time taken is = 10 + 24 = 34

Concept:
Sum of the n terms in an A.P. = n/2(2a+(n−1)d)
Where n = Number of terms,
a = First Term,
d = Common Difference.

Explanation:
We have to find the sum of 40 terms of an A.P. whose first term is 2 and common difference 4
i.e. 
n = 40, a = 2 and d = 4
Thus,
Sum = 40/2(2(2)+(40−1)(4))
⇒ Sum = 20 × (4 + (39 × 4))
⇒ Sum = 20 × (160)
⇒ Sum = 3200

Concept:
Arithmetic Progression (AP):

• The sequence of numbers where the difference of any two consecutive terms is same is called an Arithmetic Progression.
• If a be the first term, d be the common difference and n be the number of terms of an AP, then the sequence can be written as follows:
  a, a + d, a + 2d, ..., a + (n - 1)d.
• The sum of n terms of the above series is given by:

Calculation:
The given series is 5 + 9 + 13 + … + 49 which is an arithmetic progression with first term a = 5 and common difference d = 4.
Let's say that the last term 49 is the nth term.
∴ a + (n - 1)d = 49
⇒ 5 + 4(n - 1) = 49
⇒ 4(n - 1) = 44
⇒ n = 12.
And, the sum of this AP is:

Concept:
Five terms in a geometric progression:
If a G.P. has first term a and common ratio r then the five consecutive terms in the GP are of the form a/r², a/r, a, ar, ar² .

Calculation:
Let us consider a general geometric progression with common ratio r.
Assume that the five terms in the GP are 
It is given that third term is 9.
Therefore, a = 9.
Now the product of the five terms is given as follows:

But we know that a = 9.
Thus, the product is 9⁵ = 3¹⁰.

Concepts:
Let us consider sequence a₁, a₂, a₃ …. a_n is an G.P.

• Common ratio = r = 
• n^th term of the G.P. is an = arⁿ⁻¹
• Sum of n terms = s = ; where r >1
• Sum of n terms = s =; where r <1
• Sum of infinite GP = s_∞ = |r| < 1

Where a is 1^st term and r is common ratio.

Calculation:
Given: The third term of a GP is 3
Let 'a' be the first term and 'r' be the common ratio.
∴ T₃ = ar² = 3
We know that T_n = a r^n-1
So, T₁ = a, T₂ = ar, T₃ = ar², T₄ = ar³, T₅ = ar⁴
Now, Product of the first five terms = a × ar × ar² × ar³ × ar⁴ = a⁵r¹⁰ = (ar²)⁵ = 3⁵ = 243

Concepts:
Let us consider sequence a₁, a₂, a₃ …. a_n is an G.P.

• Common ratio = r = 
• n^th term of the G.P. is an = arⁿ⁻¹
• Sum of n terms = s = ; where r >1
• Sum of n terms = s =; where r <1
• Sum of infinite GP = s_∞ = |r| < 1

Calculation:
Given:
4th term of a G. P is 8 and 10th term is 27
nth  term of the G.P. is T_n = a r^n-1
∴ T₄ = a. r³ = 8      ----(1)
T₁₀ = a r⁹ = 27      ----(2)
Equation (2) ÷ (1), we get 

Concept:
The nth term of an AP is given by: T_n = a + (n - 1) × d, where a = first term and d = common difference.

Calculation:
As we know that, the nth term of an AP is given by: T_n = a + (n - 1) × d, where a = first term and d = common difference.
Let a be the first term and d is the common difference.

⇒ a_p+q = a+(p + q − 1) × d     ...1)

⇒ a_p−q = a + (p − q − 1) × d     ...2)

By adding (1) and (2), we get

⇒ a_p+q + a_p−q = 2a + 2(p−1)d = 2 × [a + (p−1)d] = 2 × a_p

Given:
T1 = 1
T2 = 3

Calculation:
T3 = T2 + T1 = 3 + 1 = 4
T4 = T3 + T2 = 4 + 3 = 7
T5 = T4 + T3 = 7 + 4 = 11
T6 = T5 + T4 = 11 + 7 = 18
T7 = T6 + T7 = 18 + 11 = 29
T8 = T7 + T6 = 29 + 18 = 47
T9 = T8 + T7 = 47 + 29 = 76
T10 = T9 + T8 = 76 + 47 = 123
T11 = T10 + T9 = 123 + 76 = 199
T1 + T2 +  T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10 + T11 = 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 + 199 = 518

Concept:
Arithmetic Progression: Sequence where the differences between every two consecutive terms are the same.
For example:- a, a +  d, a + 2d, ... are in AP.
Four terms that are in AP can be taken as
a - 3d, a - d, a + d, a + 3d
Harmonic mean (HM):  If a, b & c are in HP then,
1/a, 1/b, 1/c are in harmonic progression (AP). 

Calculation:
Given that, a, b, c, d are in H.P. then

 
Let these terms be (x - 3y), (x - y), (x + y), (x + 3y)
Hence,

Concept:
Sum of n terms of a Geometric Progression,  
 
Explanation:
Given that a_n = 2_n of the G.P.
Then, a₁ = 2
a₂ = 4
a₃ = 8
i.e. G.P. series is 2, 4, 8, 16, 32, . . .
where first term, a = 2 ;
Common ration, r = 4/2 = 8/4=...=2 ,
Number of terms, n = 6 (given in the question)

⇒ 2(64 - 1)
⇒ 2(63)
⇒ 126 

Concept:
Expansion of ex:

Calculation:

Concept:
For AP series, 
Sum of n terms  = n/2 (First term + nth term)

Calculations:
We know that, For AP series, 
the sum of n terms  = n/2 (First term + nth term)
Given, the nth term of the given series is a_n = 5n + 1.
Put n = 1, we get
a₁ = 5(1) + 1 = 6.
We know that 
sum of n terms = n/2 (First term + nth term)
⇒Sum of n terms = 
⇒Sum of n terms = 

Concept:
Let us consider sequence a₁, a₂, a₃ …. a_n is an A.P.

• Common difference “d”= a₂ – a₁ = a₃ – a₂ = …. = a_n – a_{n – 1}
• n^th term of the A.P. is given by a_n = a + (n – 1) d
• n^th term from the last is given by a_n = l – (n – 1) d 
• sum of the first n terms = S = n/2[2a + (n − 1) × d] Or sum of the first n terms = n/2(a + l)

       Where, a = First term, d = Common difference, n = number of terms and a_n = n^th term

Calculation:
Given: nth term of an A.P = a_n = 2+n/3
For first term, put n = 1
a₁ = a = (2 + 1)/3 = 3/3 = 1
For last term, put n = 97
l = (97 + 2)/3 = 99/3 = 33
We have to find the sum of first 97 terms,
S = (97/2) (1 + 33)    (∵S = n/2(a + l))
S = 97 × 17 = 1649

Concept:
If a, b, c are in A.P then 2b = a + c

Calculation:
Given:
n - 3, 4n - 2, 5n + 1 are in AP
Therefore, 2 × (4n - 2) = (n - 3) + (5n + 1)
⇒ 8n - 4 = 6n - 2
⇒ 2n = 2
∴ n = 1

Concept:
If a, b and c are in a GP then b² = ac

Calculation:
Given: The numbers - 2/7, x, - 7/2 are in GP
As we know that, if a, b and c are in GP then b2 = ac
Here, a = - 2/7, b = x and c = - 7/2
⇒ x² = (-2/7) × (-7/2) = 1
⇒ x = ± 1
Hence, correct option is 3.