Mix Examples - Arithmetic Progressions Questions in English

(A) Given the $A.P.$ is $2, 7, 12, 17, \ldots$
Here,the first term $a = 2$ and the common difference $d = 7 - 2 = 5$.
The sum of the first $n$ terms is given by the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Substituting the given values: $990 = \frac{n}{2}[2(2) + (n - 1)5]$.
$990 = \frac{n}{2}[4 + 5n - 5]$.
$990 = \frac{n}{2}[5n - 1]$.
$1980 = n(5n - 1)$.
$5n^2 - n - 1980 = 0$.
Solving the quadratic equation using the factorization method:
$5n^2 - 100n + 99n - 1980 = 0$.
$5n(n - 20) + 99(n - 20) = 0$.
$(5n + 99)(n - 20) = 0$.
This gives $n = 20$ or $n = -\frac{99}{5}$.
Since $n$ must be a natural number,we discard the negative fraction.
Therefore,$n = 20$.

(B) The formula for the sum of $n$ terms of an $A.P.$ when the first term $a$ and the last term $l$ are known is given by:
$S_n = \frac{n}{2}(a + l)$
Given:
First term $a = 5$
Last term $l = 45$
Sum of terms $S_n = 500$
Substituting the values into the formula:
$500 = \frac{n}{2}(5 + 45)$
$500 = \frac{n}{2}(50)$
$500 = 25n$
$n = \frac{500}{25}$
$n = 20$
Therefore,there are $20$ terms in the $A.P.$

(C) Given: First term $a = 5$,common difference $d = 5$,and last term $l = T_n = 95$.
Using the formula for the $n^{th}$ term of an $A.P.$: $T_n = a + (n - 1)d$.
Substituting the given values: $95 = 5 + (n - 1)5$.
Subtracting $5$ from both sides: $90 = (n - 1)5$.
Dividing by $5$: $18 = n - 1$.
Adding $1$ to both sides: $n = 19$.
Therefore,there are $19$ terms in the $A.P.$

(C) Given that the first term $a = 5$ and the $19^{th}$ term $l = T_{19} = 95$.
The number of terms $n = 19$.
The formula for the sum of $n$ terms of an $A.P.$ when the first and last terms are known is $S_{n} = \frac{n}{2}(a + l)$.
Substituting the values: $S_{19} = \frac{19}{2}(5 + 95)$.
$S_{19} = \frac{19}{2}(100)$.
$S_{19} = 19 \times 50 = 950$.
Therefore,the sum of the $19$ terms is $950$.

(B) The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$,where $a$ is the first term and $d$ is the common difference.
For $T_{30}$,we have $T_{30} = a + (30 - 1)d = a + 29d$.
For $T_{20}$,we have $T_{20} = a + (20 - 1)d = a + 19d$.
Now,calculating the difference: $T_{30} - T_{20} = (a + 29d) - (a + 19d)$.
$T_{30} - T_{20} = a - a + 29d - 19d$.
$T_{30} - T_{20} = 10d$.

(B) For an Arithmetic Progression $(A.P.)$,the $n^{th}$ term is given by $T_n = a + (n - 1)d$.
Here,the first term $a = 4$ and the common difference $d = 8 - 4 = 4$.
We need to find $T_{40} - T_{30}$.
$T_{40} = a + (40 - 1)d = a + 39d$.
$T_{30} = a + (30 - 1)d = a + 29d$.
Subtracting the two terms:
$T_{40} - T_{30} = (a + 39d) - (a + 29d) = 10d$.
Substituting the value of $d = 4$:
$T_{40} - T_{30} = 10 \times 4 = 40$.

(D) Given the Arithmetic Progression $(A.P.)$ is $3, 13, 23, 33, \ldots$
Here,the first term $a = 3$ and the common difference $d = 13 - 3 = 10$.
We need to find the term $n$ such that $T_n = T_{21} + 10$.
The formula for the $n^{th}$ term of an $A.P.$ is $T_n = a + (n - 1)d$.
Substituting the values,we get $a + (n - 1)d = (a + 20d) + 10$.
Since $d = 10$,we have $a + (n - 1)10 = a + 20(10) + 10$.
Subtracting $a$ from both sides: $(n - 1)10 = 200 + 10$.
$(n - 1)10 = 210$.
Dividing by $10$,we get $n - 1 = 21$.
Therefore,$n = 22$.

(A) The two-digit positive multiples of $7$ form an Arithmetic Progression $(A.P.)$ starting from $14$ and ending at $98$.
The sequence is: $14, 21, 28, \dots, 98$.
Here,the first term $a = 14$,the common difference $d = 7$,and the last term $T_n = 98$.
Using the formula for the $n^{th}$ term of an $A.P.$: $T_n = a + (n - 1)d$.
Substituting the values: $98 = 14 + (n - 1)7$.
Subtracting $14$ from both sides: $84 = (n - 1)7$.
Dividing by $7$: $12 = n - 1$.
Therefore,$n = 13$.
Thus,there are $13$ two-digit positive multiples of $7$.

(B) Given the sum of the first $n$ terms of an $A.P.$ is $S_{n} = 5n^{2} + 8n$.
We know that the $n^{th}$ term $T_{n}$ is given by the formula $T_{n} = S_{n} - S_{n-1}$.
First,find $S_{n-1}$ by substituting $(n-1)$ for $n$ in the expression for $S_{n}$:
$S_{n-1} = 5(n-1)^{2} + 8(n-1)$
$S_{n-1} = 5(n^{2} - 2n + 1) + 8n - 8$
$S_{n-1} = 5n^{2} - 10n + 5 + 8n - 8$
$S_{n-1} = 5n^{2} - 2n - 3$
Now,calculate $T_{n}$:
$T_{n} = (5n^{2} + 8n) - (5n^{2} - 2n - 3)$
$T_{n} = 5n^{2} + 8n - 5n^{2} + 2n + 3$
$T_{n} = 10n + 3$.

(C) Let the three consecutive terms of the $A.P.$ be $(a-d)$,$a$,and $(a+d)$.
Given that the sum of these terms is $30$:
$(a-d) + a + (a+d) = 30$
$3a = 30$
$a = 10$
Given that the product of the first and last term is $75$:
$(a-d)(a+d) = 75$
$a^2 - d^2 = 75$
Substituting $a = 10$:
$10^2 - d^2 = 75$
$100 - d^2 = 75$
$d^2 = 100 - 75 = 25$
$d = \pm 5$
Since $5$ is given as an option,the common difference is $5$.

(C) Given that $k+1, 8, k+9$ are three consecutive terms of an $A.P.$
Since the common difference $d$ is constant in an $A.P.$,we have:
$8 - (k+1) = (k+9) - 8$
$8 - k - 1 = k + 1$
$7 - k = k + 1$
$7 - 1 = k + k$
$6 = 2k$
$k = 3$

(C) The given $A.P.$ is $1, 11, 21, 31, \dots$
Here,the first term $a = 1$.
The common difference $d = 11 - 1 = 10$.
The formula for the $n^{th}$ term of an $A.P.$ is $T_n = a + (n - 1)d$.
To find the $15^{th}$ term,we substitute $n = 15$,$a = 1$,and $d = 10$ into the formula:
$T_{15} = 1 + (15 - 1) \times 10$
$T_{15} = 1 + 14 \times 10$
$T_{15} = 1 + 140$
$T_{15} = 141$
Therefore,the $15^{th}$ term is $141$.

(C) The given sequence is an arithmetic progression $(A.P.)$ where the first term $a = 13$.
The common difference $d = 26 - 13 = 13$.
The last term $T_n = 650$.
The formula for the $n^{th}$ term of an $A.P.$ is $T_n = a + (n - 1)d$.
Substituting the values: $650 = 13 + (n - 1)13$.
Dividing the entire equation by $13$: $50 = 1 + (n - 1)$.
$50 = n$,so $n = 50$.
Therefore,the number of terms in the $A.P.$ is $50$.

(D) The given arithmetic progression is $3, 8, 13, 18, \ldots$.
Here,the first term $a = 3$.
The common difference $d = 8 - 3 = 5$.
The formula for the $n^{th}$ term of an $A.P.$ is $T_{n} = a + (n - 1)d$.
Substituting the values of $a$ and $d$:
$T_{n} = 3 + (n - 1)5$
$T_{n} = 3 + 5n - 5$
$T_{n} = 5n - 2$.

(B) The sequence of even natural numbers forms an Arithmetic Progression ($A$.$P$.) given by $2, 4, 6, 8, \ldots$
Here,the first term $a = 2$,the common difference $d = 4 - 2 = 2$,and the number of terms $n = 20$.
The sum of the first $n$ terms of an $A$.$P$. is given by the formula: $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values: $S_{20} = \frac{20}{2} [2(2) + (20 - 1)2]$.
$S_{20} = 10 [4 + (19 \times 2)]$.
$S_{20} = 10 [4 + 38]$.
$S_{20} = 10 \times 42 = 420$.
Alternatively,the sum of the first $n$ even natural numbers is given by $n(n + 1)$.
For $n = 20$,sum $= 20 \times (20 + 1) = 20 \times 21 = 420$.

(D) To determine the nature of the sequence $1, 3, 6, 10, \ldots$,let us examine the differences between consecutive terms:
$3 - 1 = 2$
$6 - 3 = 3$
$10 - 6 = 4$
The differences are $2, 3, 4, \ldots$,which are not constant,so it is not an arithmetic progression.
The ratios are $3/1 = 3$,$6/3 = 2$,$10/6 = 1.66$,which are not constant,so it is not a geometric progression.
The Fibonacci sequence is defined by $F_n = F_{n-1} + F_{n-2}$,which does not match this pattern.
These numbers are known as triangular numbers,defined by the formula $T_n = \frac{n(n+1)}{2}$.
For $n=1, T_1 = 1$; for $n=2, T_2 = 3$; for $n=3, T_3 = 6$; for $n=4, T_4 = 10$. Thus,it is the sequence of triangular numbers.

(B) The given $A.P.$ is $5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, \ldots$
To find the composite numbers in this sequence,we check each term:
$5$: Prime
$7$: Prime
$9$: Composite $(3 \times 3)$
$11$: Prime
$13$: Prime
$15$: Composite $(3 \times 5)$
$17$: Prime
$19$: Prime
$21$: Composite $(3 \times 7)$
$23$: Prime
$25$: Composite $(5 \times 5)$
The composite numbers in the sequence are $9, 15, 21, 25, \ldots$
The first composite number is $9$.
The second composite number is $15$.
The third composite number is $21$.
The fourth composite number is $25$.
Therefore,the correct option is $B$.

(D) The multiples of $3$ between $1$ and $50$ form an Arithmetic Progression $(A.P.)$ as follows:
$3, 6, 9, \ldots, 48$
Here,the first term $a = 3$,the common difference $d = 3$,and the last term $T_n = 48$.
Using the formula for the $n^{th}$ term of an $A.P.$:
$T_n = a + (n - 1)d$
Substituting the values:
$48 = 3 + (n - 1)3$
$48 - 3 = (n - 1)3$
$45 = (n - 1)3$
$n - 1 = \frac{45}{3}$
$n - 1 = 15$
$n = 16$
Thus,there are $16$ multiples of $3$ between $1$ and $50$.

(B) The given arithmetic progression $(A.P.)$ is $5, 10, 15, \ldots, 200$.
Here,the first term $a = 5$.
The common difference $d = 10 - 5 = 5$.
The last term $a_n = 200$.
The formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
Substituting the values: $200 = 5 + (n - 1)5$.
$200 - 5 = (n - 1)5$.
$195 = (n - 1)5$.
$n - 1 = 195 / 5$.
$n - 1 = 39$.
$n = 39 + 1 = 40$.
Therefore,there are $40$ terms in the given $A.P.$

(A) In an $A.P.$,the common difference $d$ is defined as the difference between any two consecutive terms.
Given that $x, y, z$ are three consecutive terms of an $A.P.$
Therefore,$d = y - x$ and $d = z - y$.
Adding these two equations: $d + d = (y - x) + (z - y)$.
$2d = z - x$.
Thus,$d = \frac{z - x}{2}$.

(A) The general formula for the $n^{th}$ term of an Arithmetic Progression $(A.P.)$ is given by $a_n = a + (n - 1)d$,where $a$ is the first term,$d$ is the common difference,and $n$ is the term number.
To find the tenth term $(n = 10)$,substitute the value of $n$ into the formula:
$a_{10} = a + (10 - 1)d$
$a_{10} = a + 9d$
Therefore,the tenth term is $a + 9d$.

(D) For three terms $a, b, c$ to be in an $A.P.$,the middle term must be the arithmetic mean of the other two terms.
This is given by the formula $b = \frac{a + c}{2}$.
Here,$a = 2$,$b = x$,and $c = 20$.
Substituting these values into the formula:
$x = \frac{2 + 20}{2}$
$x = \frac{22}{2}$
$x = 11$.
Therefore,the value of $x$ is $11$.

(B) The sum of $n$ terms of an $A.P.$ when the first term $a$ and the last term $l$ are given is calculated using the formula:
$S_n = \frac{n}{2}(a + l)$
Given values are:
$a = 1$
$l = 10$
$n = 10$
Substituting these values into the formula:
$S_{10} = \frac{10}{2}(1 + 10)$
$S_{10} = 5(11)$
$S_{10} = 55$
Therefore,the correct option is $B$.

(A) The sum of $n$ terms of an $A.P.$ is given by the formula $S_n = \frac{n}{2}(a + l)$,where $a$ is the first term and $l$ is the last term.
Given values are $a = 2$,$n = 8$,and $S_8 = 72$.
Substituting these values into the formula:
$72 = \frac{8}{2}(2 + l)$
$72 = 4(2 + l)$
Dividing both sides by $4$:
$18 = 2 + l$
$l = 18 - 2$
$l = 16$
Therefore,the last term $l$ is $16$.

(A) The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.
$1$. For $A.P.$ $1, 3, 5, 7, \ldots$,$a=1, d=2$. The $6^{th}$ term is $a_6 = 1 + (6-1)2 = 1 + 10 = 11$. So,$(1-b)$.
$2$. For $A.P.$ $3, 6, 9, 12, \ldots$,$a=3, d=3$. The $11^{th}$ term is $a_{11} = 3 + (11-1)3 = 3 + 30 = 33$. So,$(2-c)$.
$3$. For $A.P.$ $4, 6, 8, 10, \ldots$,$a=4, d=2$. The $16^{th}$ term is $a_{16} = 4 + (16-1)2 = 4 + 30 = 34$. So,$(3-d)$.
$4$. For $A.P.$ $5, 10, 15, 20, \ldots$,$a=5, d=5$. The $21^{st}$ term is $a_{21} = 5 + (21-1)5 = 5 + 100 = 105$. So,$(4-a)$.
Thus,the correct matching is $(1-b), (2-c), (3-d), (4-a)$.

(A) The formula for the $n^{th}$ term of the $A.P.$ is given as $T_{n} = 2n - 1$.
To find the $10^{th}$ term,we substitute $n = 10$ into the given formula.
$T_{10} = 2(10) - 1$
$T_{10} = 20 - 1$
$T_{10} = 19$
Therefore,the $10^{th}$ term of the $A.P.$ is $19$.