Series completion Questions in English

(D) The pattern of the series is as follows:
$647 - 563 = +84$
$479 - 647 = -168$ (which is $-(84 \times 2)$)
$815 - 479 = +336$ (which is $+(84 \times 4)$ or $+(84 \times 2^2)$)
Following this pattern,the next term should involve subtracting $84 \times 2^3 = 84 \times 8 = 672$.
Missing number $= 815 - 672 = 143$.

(B) The given sequence is a combination of two series:
$I.$ $11, 17, 23, (\ldots)$
$II.$ $12, 18, 24$
The pattern in both series $I$ and $II$ is an addition of $6$ to the previous term.
For series $I$: $11 + 6 = 17$,$17 + 6 = 23$,$23 + 6 = 29$.
Therefore,the missing number is $29$.

(C) Analyze the pattern of the given series: $225, 336, 447, (...), 669, 7710$.
$1$. Observe the first two digits of each number: $22, 33, 44, 55, 66, 77$. This forms an arithmetic progression where each term increases by $11$.
$2$. Observe the third digit (or remaining part) of each number: $5, 6, 7, 8, 9, 10$. This forms a simple arithmetic progression where each term increases by $1$.
$3$. Combining these patterns,the missing number should have the first two digits as $55$ and the third digit as $8$.
Therefore,the missing number is $558$.

(B) Observe the relationship between consecutive terms:
$840 \div 5 = 168$
$168 \div 4 = 42$
$42 \div 3 = 14$
$14 \div 2 = 7$
Following the pattern of dividing by decreasing integers $(5, 4, 3, 2)$,the next operation is to divide by $1$.
Missing number $= 7 \div 1 = 7$.

(A) The given sequence is a combination of two alternating series:
Series $I$: $5, 7, 10, 14, \ldots$
Pattern: $+2, +3, +4, \ldots$
Series $II$: $6, 8, 11, (\ldots)$
Pattern: $+2, +3, +4, \ldots$
Following the pattern for Series $II$,the next term is $11 + 4 = 15$.

(A) The given sequence is a combination of two alternating series:
Series $I$: $0, 3, 8, 15, 24, (\ldots)$
Pattern for Series $I$: The differences are $+3, +5, +7, +9$. The next difference should be $+11$.
So,the next term is $24 + 11 = 35$.
Series $II$: $2, 5, 10, 17, 26$
Pattern for Series $II$: The differences are $+3, +5, +7, +9$.
Therefore,the missing number is $35$.

(B) The given sequence is a combination of three interleaved series:
Series $I$: $0, 3, 6, \dots$ (Pattern: $+3$)
Series $II$: $4, 7, (\dots)$ (Pattern: $+3$)
Series $III$: $6, 9, 12$ (Pattern: $+3$)
Looking at Series $II$,the terms are $4, 7, x$. Since the common difference is $+3$,the missing number is $7 + 3 = 10$.
Therefore,the correct option is $B$.

(A) The given sequence is a combination of two alternating series:
Series $I$: $1, 3, 6, 10, (\ldots)$
Pattern: $+2, +3, +4, +5, \ldots$
So,the next term is $10 + 5 = 15$.
Series $II$: $1, 9, 36, 100, 225$
Pattern: $1^2, 3^2, 6^2, 10^2, 15^2$
Thus,the missing number is $15$.

(B) The given sequence is a combination of three interleaved series:
$I.$ $1^{st}, 4^{th}, 7^{th}, 10^{th}, 13^{th}$ terms: $2, 4, 6, 8, (\ldots)$
$II.$ $2^{nd}, 5^{th}, 8^{th}, 11^{th}$ terms: $1, 4, 7, 10$
$III.$ $3^{rd}, 6^{th}, 9^{th}, 12^{th}$ terms: $2, 5, 8, 11$
In series $I$,the pattern is an arithmetic progression with a common difference of $2$. The next term after $8$ is $8 + 2 = 10$.
Thus,the missing number is $10$.

(A) The given series is $4, 23, 60, 121, \ldots$
Observe the pattern of the terms:
$2^{3} - 4 = 8 - 4 = 4$
$3^{3} - 4 = 27 - 4 = 23$
$4^{3} - 4 = 64 - 4 = 60$
$5^{3} - 4 = 125 - 4 = 121$
The pattern follows the rule $n^{3} - 4$,where $n$ starts from $2$.
Therefore,the next term is $6^{3} - 4 = 216 - 4 = 212$.

(A) The given series is $1, 4, 2, 8, 6, 24, 22, 88, (\ldots)$.
Observe the pattern between consecutive terms:
$1 \times 4 = 4$
$4 - 2 = 2$
$2 \times 4 = 8$
$8 - 2 = 6$
$6 \times 4 = 24$
$24 - 2 = 22$
$22 \times 4 = 88$
The pattern follows the sequence of operations: $\times 4, -2, \times 4, -2, \times 4, -2, \times 4, \ldots$
Following this pattern,the next operation should be $-2$.
Therefore,the missing number is $88 - 2 = 86$.

(C) The given sequence is a combination of two alternating series:
Series $I: 13, 24, 35, 46, 57$
Pattern: Each term increases by $11$ $(13+11=24, 24+11=35, 35+11=46, 46+11=57)$.
Series $II: 32, 43, (\ldots), 65, 76$
Pattern: Each term increases by $11$ $(32+11=43, 43+11=54, 54+11=65, 65+11=76)$.
Therefore,the missing number is $43+11=54$.

(B) The given sequence is a combination of two alternating series:
$I.$ Odd-numbered terms: $3, 7, 13, 21, 31, \ldots$
The differences are: $(7-3)=4, (13-7)=6, (21-13)=8, (31-21)=10$. The next difference should be $12$. Thus,the next term in this series is $31+12=43$.
$II.$ Even-numbered terms: $4, 7, 13, 22, 44, \ldots$
The differences are: $(7-4)=3, (13-7)=6, (22-13)=9, (44-22)=22$. (Note: This series follows a different progression).
Since the missing term is the $11^{th}$ term (an odd-numbered position),it follows the pattern of series $I$. Therefore,the missing term is $31+12=43$.

(D) The given series is $2, 6, 12, 20, 30, 42, 56, \dots$
We can observe the pattern as follows:
$1 \times 2 = 2$
$2 \times 3 = 6$
$3 \times 4 = 12$
$4 \times 5 = 20$
$5 \times 6 = 30$
$6 \times 7 = 42$
$7 \times 8 = 56$
Following this pattern,the next term is $8 \times 9 = 72$.

(A) The given sequence is a combination of three alternating series:
$I.$ $1^{st}, 4^{th}, 7^{th}, 10^{th}$ terms: $8, 7, 6, (\dots)$
$II.$ $2^{nd}, 5^{th}, 8^{th}, 11^{th}$ terms: $9, 10, 11, 12$
$III.$ $3^{rd}, 6^{th}, 9^{th}$ terms: $8, 9, 10$
In series $I$,the pattern is a decrease of $1$ at each step $(8-1=7, 7-1=6)$.
Therefore,the missing number is $6-1 = 5$.

(A) The series follows a pattern of multiplication in groups of three: $(a, b, c)$ where $c \times a = b$.
$1$. Group $1$: $(30, 3, 90) \rightarrow 30 \times 3 = 90$
$2$. Group $2$: $(6, 30, 180) \rightarrow 6 \times 30 = 180$
$3$. Group $3$: $(2, 6, 12) \rightarrow 2 \times 6 = 12$
$4$. Group $4$: $(25, 2, 50) \rightarrow 25 \times 2 = 50$
$5$. Group $5$: $(4, 25, 100) \rightarrow 4 \times 25 = 100$
$6$. Group $6$: $(50, 4, 200) \rightarrow 50 \times 4 = 200$
$7$. Group $7$: $(x, 50, 3) \rightarrow$ Wait,looking at the sequence $90, 180, 12, 50, 100, 200, (\ldots), 3, 50, 4, 25, 2, 6, 30, 3$.
Re-evaluating the pattern: The sequence is $90, 180, 12$ (where $12 \times 15 = 180$ is not it). Let's look at the triplets: $(90, 180, 12)$ is not standard. Let's look at the sequence again: $90, 180, 12$ is $90 \times 2 = 180$,$180 / 15 = 12$. $50, 100, 200$ is $50 \times 2 = 100, 100 \times 2 = 200$. The missing number is $150$ because $3 \times 50 = 150$.

(A) The given sequence is a combination of two alternating series:
Series $I$: $11, (...), 1001, 10001$
Series $II$: $10, 100, 1000$
In Series $I$,the pattern involves adding an extra zero between the two $1$s as the number progresses $(11 \rightarrow 101 \rightarrow 1001 \rightarrow 10001)$.
Therefore,the missing number is $101$.

(D) Let us analyze the pattern of the series:
$1$. The first term is $123456147$.
$2$. The second term is $12345614$ (the last digit '$7$' is removed).
$3$. The third term is $2345614$ (the first digit '$1$' is removed from the second term).
$4$. The fourth term is $234561$ (the last digit '$4$' is removed from the third term).
$5$. Following this alternating pattern of removing the last digit and then the first digit,the next step is to remove the first digit '$2$' from the fourth term.
Therefore,the next term is $34561$.

(B) The given series is $2, 6, 12, 20, 30, ?$.
Observe the differences between consecutive terms:
$6 - 2 = 4$
$12 - 6 = 6$
$20 - 12 = 8$
$30 - 20 = 10$
The differences are $4, 6, 8, 10$,which form an arithmetic progression with a common difference of $2$.
The next difference should be $10 + 2 = 12$.
Therefore,the next term is $30 + 12 = 42$.
Alternatively,the series follows the pattern $n^2 + n$ (or $n(n+1)$):
$1^2 + 1 = 2$
$2^2 + 2 = 6$
$3^2 + 3 = 12$
$4^2 + 4 = 20$
$5^2 + 5 = 30$
$6^2 + 6 = 42$
Thus,the correct option is $B$.

(B) The given series is $2, 6, 12, 20, 30, ?$.
Let's analyze the differences between consecutive terms:
$6 - 2 = 4$
$12 - 6 = 6$
$20 - 12 = 8$
$30 - 20 = 10$
The differences are $4, 6, 8, 10$,which form an arithmetic progression with a common difference of $2$.
The next difference should be $10 + 2 = 12$.
Therefore,the next term is $30 + 12 = 42$.
Alternatively,the series follows the pattern $n^2 + n$:
$1^2 + 1 = 2$
$2^2 + 2 = 6$
$3^2 + 3 = 12$
$4^2 + 4 = 20$
$5^2 + 5 = 30$
$6^2 + 6 = 42$.

(B) The given series is $2, 6, 12, 20, 30, ?$.
We can observe the pattern by finding the differences between consecutive terms:
$6 - 2 = 4$
$12 - 6 = 6$
$20 - 12 = 8$
$30 - 20 = 10$
The differences are $4, 6, 8, 10$,which form an arithmetic progression with a common difference of $2$.
The next difference should be $10 + 2 = 12$.
Therefore,the next term is $30 + 12 = 42$.
Alternatively,the series follows the pattern $n^2 + n$ or $n(n+1)$:
$1^2 + 1 = 2$
$2^2 + 2 = 6$
$3^2 + 3 = 12$
$4^2 + 4 = 20$
$5^2 + 5 = 30$
$6^2 + 6 = 42$.
Thus,the correct option is $B$.

(C) The given series consists of pairs of consecutive prime numbers.
- The first pair is $(2, 3)$.
- The second pair is $(3, 5)$.
- The third pair is $(5, 7)$.
- The fourth pair is $(7, 11)$.
- The fifth pair is $(11, 13)$.
Following this pattern,the next pair of consecutive prime numbers after $13$ is $(13, 17)$.

(B) The given series is an arithmetic progression with the first term $a = 10$ and common difference $d = 7$.
The general term of the series is given by $T_n = a + (n - 1)d = 10 + (n - 1)7 = 7n + 3$.
This means any number in the series must satisfy the form $7n + 3$,which is equivalent to saying that when the number is divided by $7$,the remainder is $3$.
Let us check the options:
$A) 48 = 7 \times 6 + 6$ (Remainder $6$)
$B) 346 = 7 \times 49 + 3$ (Remainder $3$)
$C) 574 = 7 \times 82 + 0$ (Remainder $0$)
$D) 1003 = 7 \times 143 + 2$ (Remainder $2$)
Thus,$346$ is the only number that belongs to the series.

(A) The given series is $1^3, 2^3, 3^3, 4^3, 5^3, \ldots$ which represents the cubes of natural numbers.
$1^3 = 1$
$2^3 = 8$
$3^3 = 27$
$4^3 = 64$
$5^3 = 125$
Checking the options:
$A) 256 = 16^2$ (not a perfect cube)
$B) 512 = 8^3$
$C) 729 = 9^3$
$D) 1000 = 10^3$
Therefore,$256$ is not a number in the series.

(C) The given series is $3, 9, 15, . . .$
Here,the difference between consecutive terms is constant: $9 - 3 = 6$ and $15 - 9 = 6$.
Thus,this is an Arithmetic Progression $(A.P.)$ where the first term $a = 3$ and the common difference $d = 6$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
For the $21^{st}$ term $(n = 21)$:
$a_{21} = 3 + (21 - 1) \times 6$
$a_{21} = 3 + 20 \times 6$
$a_{21} = 3 + 120 = 123$.

(B) The given series is $2, 6, 18, 54, \ldots$
Here,each term is obtained by multiplying the previous term by $3$.
This is a Geometric Progression $(G.P.)$ where the first term $a = 2$ and the common ratio $r = 3$.
The formula for the $n^{th}$ term of a $G.P.$ is $T_n = a \cdot r^{n-1}$.
For the $8^{th}$ term $(n = 8)$:
$T_8 = 2 \cdot 3^{8-1} = 2 \cdot 3^7$.
Calculating $3^7 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2187$.
Therefore,$T_8 = 2 \times 2187 = 4374$.

(C) The given series is $5, 8, 11, 14, \ldots$
This is an Arithmetic Progression $(A.P.)$ where the first term $a = 5$ and the common difference $d = 8 - 5 = 3$.
We need to find the term $n$ such that the $n^{th}$ term $a_n = 320$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values: $320 = 5 + (n - 1) \times 3$.
$320 - 5 = (n - 1) \times 3$.
$315 = (n - 1) \times 3$.
$n - 1 = \frac{315}{3} = 105$.
$n = 105 + 1 = 106$.
Therefore,the $106^{th}$ term of the series is $320$.

(B) The given series is $5, 10, 20, 40, \ldots$
This is a Geometric Progression $(G.P.)$ where the first term $a = 5$ and the common ratio $r = \frac{10}{5} = 2$.
The $n^{th}$ term of a $G.P.$ is given by the formula $a_n = a \cdot r^{n-1}$.
We need to find $n$ such that $a_n = 1280$.
Substituting the values,we get $5 \cdot 2^{n-1} = 1280$.
Dividing both sides by $5$,we get $2^{n-1} = \frac{1280}{5} = 256$.
Since $256 = 2^8$,we have $2^{n-1} = 2^8$.
Equating the exponents,$n - 1 = 8$,which gives $n = 9$.
Therefore,$1280$ is the $9^{th}$ term of the series.

(D) The given series follows the pattern $(n^{3} - 1)$ for $n = 2, 3, 4, 5, 6, 7, 8, 9$.
Calculating the terms:
$2^{3} - 1 = 8 - 1 = 7$
$3^{3} - 1 = 27 - 1 = 26$
$4^{3} - 1 = 64 - 1 = 63$
$5^{3} - 1 = 125 - 1 = 124$
$6^{3} - 1 = 216 - 1 = 215$
$7^{3} - 1 = 343 - 1 = 342$
$8^{3} - 1 = 512 - 1 = 511$
$9^{3} - 1 = 729 - 1 = 728$
Comparing this with the given series $7, 28, 63, 124, 215, 342, 611$,we see that $28$ is incorrect (it should be $26$) and $611$ is incorrect (it should be $511$). However,looking at the options provided,$611$ is the most significant outlier in the sequence pattern.

(C) The given series is $3, 8, 15, 24, 34, 48, 63$.
Let us observe the pattern of differences between consecutive terms:
$8 - 3 = 5$
$15 - 8 = 7$
$24 - 15 = 9$
$35 - 24 = 11$
$48 - 35 = 13$
$63 - 48 = 15$
The differences should be consecutive odd numbers: $5, 7, 9, 11, 13, 15$.
In the given series,the difference between $24$ and $34$ is $10$,and between $34$ and $48$ is $14$.
Therefore,$34$ is the wrong number and should be replaced by $35$ to maintain the pattern of odd differences.

(C) Analyze the pattern of the series:
$24 + 3 = 27$
$27 + 3 = 30$
$30 + 3 = 33$
$33 + 3 = 36$
In the given series,the term $31$ is incorrect because the pattern requires adding $3$ to each preceding term. Replacing $31$ with $30$ makes the series consistent $(24, 27, 30, 33, 36)$. Therefore,$31$ is the wrong term.

(A) The given series is $196, 169, 144, 121, 80$.
Observe the pattern of the terms:
$196 = (14)^2$
$169 = (13)^2$
$144 = (12)^2$
$121 = (11)^2$
The next term in the sequence should be $(10)^2 = 100$.
However,the given term is $80$.
Therefore,$80$ is the wrong term in the series.

(D) The given series is $3, 5, 7, 9, 11, 13$.
These numbers are intended to be consecutive prime numbers.
$A$ prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.
The prime numbers starting from $3$ are $3, 5, 7, 11, 13, 17, \dots$
In the given series,$9$ is a composite number because $9 = 3 \times 3$.
Therefore,$9$ is the wrong term in the series.

(C) The pattern in the series is that each term is obtained by adding $22$ to the previous term.
$121 + 22 = 143$
$143 + 22 = 165$
$165 + 22 = 187$
$187 + 22 = 209$
Comparing this with the given series $(121, 143, 165, 186, 209)$,we can see that $186$ is the incorrect term. It should be $187$.

(D) The given series follows a pattern where each term is obtained by multiplying the preceding term by $2$.
$1 \times 2 = 2$
$2 \times 2 = 4$
$4 \times 2 = 8$
$8 \times 2 = 16$
$16 \times 2 = 32$
$32 \times 2 = 64$
$64 \times 2 = 128$
In the given series,the last term is $96$,which does not follow the pattern. Therefore,$96$ is the wrong term and should be replaced by $128$.

(D) The pattern in the series is that each term is $2$ less than twice the preceding term.
Let us check the terms:
$8 \times 2 - 2 = 14$
$14 \times 2 - 2 = 26$
$26 \times 2 - 2 = 50$ (but the given term is $48$)
$50 \times 2 - 2 = 98$
$98 \times 2 - 2 = 194$
$194 \times 2 - 2 = 386$
Therefore,the term $48$ is incorrect and should be replaced by $50$.

(D) Let us analyze the differences between consecutive terms:
$13 - 8 = 5$
$21 - 13 = 8$
$32 - 21 = 11$
$47 - 32 = 15$
$63 - 47 = 16$
$83 - 63 = 20$
The pattern of differences is $5, 8, 11, 14, 17, 20$ (an arithmetic progression with a common difference of $3$).
Calculating the correct terms based on this pattern:
$8 + 5 = 13$
$13 + 8 = 21$
$21 + 11 = 32$
$32 + 14 = 46$
$46 + 17 = 63$
$63 + 20 = 83$
Thus,$47$ is the wrong term as it should be $46$.

(C) The given series follows a pattern of groups of three: $(n, n^2, n^3)$.
Group $1$: $3, 3^2, 3^3$ which is $3, 9, 27$.
Group $2$: $4, 4^2, 4^3$ which is $4, 16, 64$.
Group $3$: $5, 5^2, 5^3$ which is $5, 25, 125$.
Comparing this with the given series $3, 10, 27, 4, 16, 64, 5, 25, 125$,we can see that $10$ is the wrong term,as it should be $9$.

(C) Let us analyze the pattern of the series from right to left:
$2 \times 2 + 4 = 8$
$8 \times 2 + 4 = 20$
$20 \times 2 + 4 = 44$
$44 \times 2 + 4 = 92$
$92 \times 2 + 4 = 188$
$188 \times 2 + 4 = 380$
Comparing this with the given series $380, 188, 92, 48, 20, 8, 2$,we see that $48$ is the wrong term,as it should be $44$.

(C) The given series is $1, 3, 7, 15, 27, 63, 127$.
Let us analyze the differences between consecutive terms:
$3 - 1 = 2 = 2^1$
$7 - 3 = 4 = 2^2$
$15 - 7 = 8 = 2^3$
$27 - 15 = 12$
$63 - 27 = 36$
$127 - 63 = 64 = 2^6$
The pattern of differences should be powers of $2$ $(2^1, 2^2, 2^3, 2^4, 2^5, 2^6)$.
Following this pattern,the difference after $15$ should be $2^4 = 16$.
So,the next term should be $15 + 16 = 31$.
Then,the next difference should be $2^5 = 32$,and $31 + 32 = 63$,which matches the next term.
Therefore,$27$ is the wrong term in the series.

(C) The given series is $5, 10, 17, 24, 37$.
Let us analyze the differences between consecutive terms:
$10 - 5 = 5$
$17 - 10 = 7$
$24 - 17 = 7$
$37 - 24 = 13$
Observing the pattern of differences: $5, 7, 9, 11, \dots$
The differences should be consecutive odd numbers starting from $5$.
If we replace $24$ with $26$,the series becomes $5, 10, 17, 26, 37$.
The differences would then be:
$10 - 5 = 5$
$17 - 10 = 7$
$26 - 17 = 9$
$37 - 26 = 11$
This follows the pattern of consecutive odd numbers.
Therefore,$24$ is the wrong term.

(D) Let us analyze the pattern of the series:
$1 \times 2 + 1 = 3$
$3 \times 3 + 1 = 10$
$10 \times 2 + 1 = 21$
$21 \times 3 + 1 = 64$
$64 \times 2 + 1 = 129$
$129 \times 3 + 1 = 388$
$388 \times 2 + 1 = 777$ (Note: The series logic follows $\times 2+1, \times 3+1$ alternately).
Comparing this with the given series $1, 3, 10, 21, 64, 129, 256, 778$,we see that $256$ is the incorrect term as the expected term after $129$ is $388$.

(B) Let us analyze the differences between consecutive terms:
$16 - 15 = 1 = 1^2$
$22 - 16 = 6 \neq 2^2$
$29 - 22 = 7 \neq 3^2$
Let us look at the pattern of adding squares to the previous term:
$15 + 1^2 = 16$
$16 + 2^2 = 16 + 4 = 20$
$20 + 3^2 = 20 + 9 = 29$
$29 + 4^2 = 29 + 16 = 45$
$45 + 5^2 = 45 + 25 = 70$
Comparing this to the given series $15, 16, 22, 29, 45, 70$,we see that $22$ is the wrong term and should be $20$.

(C) The pattern followed in the series is:
$6 \times 2 + 2 = 14$
$14 \times 2 + 2 = 30$
$30 \times 2 + 2 = 62$
$62 \times 2 + 2 = 126$
Comparing this with the given series $6, 14, 30, 64, 126$,we can see that $64$ is the wrong term and should be $62$.

(D) The pattern of the series is that each term is $4$ less than $3$ times the preceding term.
Let's check the terms:
$10 \times 3 - 4 = 26$
$26 \times 3 - 4 = 74$
$74 \times 3 - 4 = 218$
$218 \times 3 - 4 = 650$
$650 \times 3 - 4 = 1946$
$1946 \times 3 - 4 = 5834$
Comparing this with the given series,the term $654$ is incorrect and should be $650$.

(B) The pattern of the series is that each term is obtained by multiplying the previous term by $2$ and adding $1$ $(x_{n+1} = 2x_n + 1)$.
Let's check the terms:
$3 \times 2 + 1 = 7$
$7 \times 2 + 1 = 15$
$15 \times 2 + 1 = 31$ (not $39$)
$31 \times 2 + 1 = 63$
$63 \times 2 + 1 = 127$
$127 \times 2 + 1 = 255$
$255 \times 2 + 1 = 511$
Therefore,$39$ is the wrong term in the series and should be replaced by $31$.

(B) The pattern followed in the series is: subtract $3$ from the current term and then divide by $2$ to get the next term.
$1.$ $(445 - 3) / 2 = 442 / 2 = 221$
$2.$ $(221 - 3) / 2 = 218 / 2 = 109$
$3.$ $(109 - 3) / 2 = 106 / 2 = 53$
$4.$ $(53 - 3) / 2 = 50 / 2 = 25$
$5.$ $(25 - 3) / 2 = 22 / 2 = 11$
$6.$ $(11 - 3) / 2 = 8 / 2 = 4$
Comparing this with the given series,the term $46$ is incorrect and should be $53$.

(D) Let us analyze the pattern of the digits in each number:
$1$. The first digits are $1, 2, 3, 4, 5$. This follows a consistent pattern of increasing by $1$.
$2$. The second digits are $2, 3, 4, 5, 6$. This follows a consistent pattern of increasing by $1$.
$3$. The third digits are $3, 4, 5, 6, 7$. Following this pattern,the third digit of the fifth number should be $7$.
$4$. The last digit in each number is $6$.
Therefore,the fifth number should be $5676$ instead of $5686$. Thus,$5686$ is the wrong term.

(C) The pattern in the series is alternating multiplication by $2$ and $4$:
$5 \times 2 = 10$
$10 \times 4 = 40$
$40 \times 2 = 80$
$80 \times 4 = 320$
$320 \times 2 = 640$
$640 \times 4 = 2560$
Comparing this with the given series,the term $550$ is incorrect and should be $640$.

(B) The given series is a combination of two alternating series:
Series $I$: $3, 8, 13, 18, 23$ (Pattern: $+5$)
Series $II$: $2, 9, 22, 32, 42$ (Pattern: $+10$)
In Series $I$,the terms follow the rule $a_n = 3 + (n-1)5$,which is correct.
In Series $II$,the terms should be $2, 12, 22, 32, 42$ (adding $10$ to each previous term).
Comparing the given Series $II$ $(2, 9, 22, 32, 42)$ with the correct series $(2, 12, 22, 32, 42)$,we see that $9$ is the wrong term and should be $12$.