EXAMPLES

1. Identify the first term of the given AP.
The sequence starts as $3, 1, -1, -3, \dots$; therefore the first term $a = 3$.

2. Find the common difference $d$ using any two consecutive terms.
Using the first two terms: \[d = 1 - 3 = -2.\]
(Verification with other consecutive terms: $-1 - 1 = -2$, $-3 - (-1) = -2$, confirming the constant difference.)

3. Hence, the required values are: \[a = 3, \quad d = -2.\]

1. Check each list for a constant common difference
- Compute successive differences $d_i = a_{i+1} - a_i$.
- If all $d_i$ are equal, the list is an AP; otherwise it is not.

(i) 4, 10, 16, 22, …
\[ d_1 = 10-4 = 6,\; d_2 = 16-10 = 6,\; d_3 = 22-16 = 6 \]
All differences are $6$, so the sequence is an AP with common difference $d = 6$.
Next two terms:
\[ a_5 = 22 + 6 = 28,\; a_6 = 28 + 6 = 34 \]

(ii) 1, –1, –3, –5, …
\[ d_1 = (-1)-1 = -2,\; d_2 = (-3)-(-1) = -2,\; d_3 = (-5)-(-3) = -2 \]
All differences are $-2$, so the sequence is an AP with $d = -2$.
Next two terms:
\[ a_5 = -5 + (-2) = -7,\; a_6 = -7 + (-2) = -9 \]

(iii) –2, 2, –2, 2, –2, …
\[ d_1 = 2-(-2) = 4,\; d_2 = (-2)-2 = -4,\; d_3 = 2-(-2) = 4 \]
The differences alternate between $4$ and $-4$; they are not constant. Hence this list does not form an AP.

(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, …
\[ d_1 = 1-1 = 0,\; d_2 = 1-1 = 0,\; d_3 = 2-1 = 1,\; d_4 = 2-2 = 0,\; d_5 = 2-2 = 0,\; d_6 = 3-2 = 1 \]
The successive differences are $0,0,1,0,0,1,\dots$ – not a single constant value. Therefore this list does not form an AP.

2. Summarize the results
- (i) Forms an AP; next two terms are $28$ and $34$.
- (ii) Forms an AP; next two terms are $-7$ and $-9$.
- (iii) Does not form an AP.
- (iv) Does not form an AP.

1. First term: From the given AP, the first term \(a = 2\).

2. Common difference: \(d = 7-2 = 5\) (or \(12-7 = 5\)).

3. Nth‑term formula: For an AP, the \(n^{th}\) term is given by
$$a_n = a + (n-1)d.$$

4. Substitute \(n = 10\):
$$a_{10} = 2 + (10-1)\times 5$$
$$a_{10} = 2 + 9\times 5$$
$$a_{10} = 2 + 45$$
$$a_{10} = 47.$$

1. Identify the first term and common difference:
\[ a = 21, \quad d = 18-21 = -3 \]

2. Write the general term of the AP:
\[ a_n = a + (n-1)d = 21 + (n-1)(-3) \]

3. Find the term that equals \(-81\):
\[ -81 = 21 - 3(n-1) \]
\[ -81 - 21 = -3(n-1) \]
\[ -102 = -3(n-1) \]
Divide both sides by \(-3\):
\[ 34 = n-1 \]
\[ n = 35 \]
Hence, the 35th term of the AP is \(-81\).

4. Check whether any term is \(0\):
Set \(a_n = 0\):
\[ 0 = 21 - 3(n-1) \]
\[ -3(n-1) = -21 \]
Divide by \(-3\):
\[ n-1 = 7 \]
\[ n = 8 \]
Since \(n = 8\) is a positive integer, the 8th term of the AP is \(0\).

5. Reason: The common difference is \(-3\); each successive term reduces by 3. After 7 steps from the first term (21), the value becomes \(21 - 7\times3 = 0\). Thus a zero term indeed occurs.

6. Answer summary:
- The term that equals \(-81\) is the 35th term.
- Yes, a term equals \(0\); it is the 8th term.

1. Let $a$ be the first term and $d$ be the common difference of the AP.
2. Write the expression for the 3rd term using the AP formula:
$$a_3 = a + 2d = 5 \quad\text{(Equation 1)}$$
3. Write the expression for the 7th term:
$$a_7 = a + 6d = 9 \quad\text{(Equation 2)}$$
4. Subtract Equation 1 from Equation 2 to eliminate $a$:
$$(a + 6d) - (a + 2d) = 9 - 5$$
$$4d = 4$$
$$d = 1$$
5. Substitute $d = 1$ back into Equation 1 to find $a$:
$$a + 2(1) = 5 \Rightarrow a + 2 = 5 \Rightarrow a = 3$$
6. Hence, the required AP is:
$$\boxed{3,\;4,\;5,\;6,\;7,\;8,\;9,\;\dots}$$
(First term $a = 3$, common difference $d = 1$).

1. Identify the first term and common difference of the given AP.
\[ a = 5 \]
The successive terms increase by \(6\) (\(11-5 = 6\)), so \( d = 6 \).

2. Write the general (nth) term of the AP.
\[ a_n = a + (n-1)d = 5 + (n-1)\times 6 \]

3. Set the given number \(301\) equal to the nth term and solve for \(n\).
\[ 301 = 5 + (n-1)\times 6 \]
\[ 301 - 5 = (n-1)\times 6 \]
\[ 296 = 6(n-1) \]
\[ n-1 = \frac{296}{6} = 49\frac{1}{3} \]

4. Since \(n-1\) is not an integer, \(n\) is not a natural number. Hence \(301\) cannot be expressed as \(a_n\) for any integer \(n\ge 1\).

5. Alternatively, check the divisibility condition:
\[ 301 - 5 = 296 \]
\[ 296 \div 6 = 49\frac{1}{3} \] (not an integer). Therefore \(301\) is not a term of the AP.

6. Conclude: 301 is not a term of the given list.

1. Identify the range of two‑digit numbers: \(10 \leq N \leq 99\).

2. Find the smallest two‑digit number divisible by 3:
\[10 \equiv 1 \pmod{3},\; 11 \equiv 2 \pmod{3},\; 12 \equiv 0 \pmod{3}\]
Hence the first term \(a = 12\).

3. Find the largest two‑digit number divisible by 3:
\[99 \equiv 0 \pmod{3}\]
Hence the last term \(l = 99\).

4. Recognize the AP: The numbers \(12, 15, 18, \dots , 99\) constitute an AP with common difference \(d = 3\).

5. Use the AP term‑count formula:
\[n = \frac{l - a}{d} + 1 = \frac{99 - 12}{3} + 1 = \frac{87}{3} + 1 = 29 + 1 = 30\]

6. Conclusion: There are \(30\) two‑digit numbers that are divisible by 3.

1. Identify the first term and common difference:
$$a = 10, \quad d = 7-10 = -3.$$\
2. Let the total number of terms be $n$. The last term $a_n$ is given as $-62$.
$$a_n = a + (n-1)d = -62.$$\
3. Substitute $a$ and $d$:
$$10 + (n-1)(-3) = -62$$
$$-3(n-1) = -72$$
$$n-1 = 24 \Rightarrow n = 25.$$\
4. The 11th term from the last term means we move 10 positions towards the first term. Hence the required term is the $(n-10)^{\text{th}}$ term:
$$\text{Required term} = a_{n-10} = a_{25-10}=a_{15}.$$\
5. Compute the 15th term using the nth term formula:
$$a_{15} = a + (15-1)d = 10 + 14(-3) = 10 - 42 = -32.$$\
6. Therefore, the 11th term from the last term is $-32$.

1. Interest for the first year
\[ I_1 = P \times \frac{R}{100} = 1000 \times \frac{8}{100} = 80 \text{ rupees} \]

2. Interest for subsequent years
Since the interest is *simple*, the same principal $P = 1000$ rupees is used every year. Hence
\[ I_2 = I_3 = \dots = I_n = 80 \text{ rupees} \]
The sequence of yearly interests is \[ 80,\,80,\,80,\,\dots \]

3. Check if the sequence is an AP
An AP has the form $a, a+d, a+2d, \dots$ where $a$ is the first term and $d$ the common difference. Here $a = 80$ and each successive term differs by $0$, i.e., $d = 0$. Therefore the interests do form an AP (a constant AP).

4. Use the AP formula to find the interest at the end of the 30th year
The $n^{\text{th}}$ term of an AP is \[ a_n = a + (n-1)d \]
Substituting $a = 80$, $d = 0$, and $n = 30$ gives
\[ a_{30} = 80 + (30-1)\times 0 = 80 \text{ rupees} \]
Hence the interest earned in the 30th year is Rs. 80.

5. (Optional) Total interest after 30 years
Total interest $= 30 \times 80 = 2400$ rupees.

1. Identify the AP:
- First term \(a_1 = 23\).
- Second term \(21\) ⇒ common difference \(d = 21-23 = -2\).
2. The last term given is \(a_n = 5\).
3. Use the nth‑term formula of an AP:
$$a_n = a_1 + (n-1)d$$
Substitute the known values:
$$5 = 23 + (n-1)(-2)$$
4. Solve for \(n\):
$$5 - 23 = (n-1)(-2)$$
$$-18 = -2(n-1)$$
$$n-1 = \frac{-18}{-2} = 9$$
$$n = 9 + 1 = 10$$
5. Hence, the flower bed contains 10 rows of rose plants.

1. Identify the first term \(a\) and the common difference \(d\).
\[a = 8\]
\[d = 3-8 = -5\]

2. Number of terms required \(n = 22\).

3. Find the \(22^{nd}\) term (last term) \(l\) using \(l = a + (n-1)d\).
\[l = 8 + (22-1)(-5) = 8 + 21(-5) = 8 - 105 = -97\]

4. Apply the sum formula \(S_n = \frac{n}{2}(a + l)\).
\[S_{22} = \frac{22}{2}\bigl(8 + (-97)\bigr) = 11 \times (-89)\]

5. Calculate the product.
\[S_{22} = -979\]

Hence, the sum of the first 22 terms of the given AP is \(-979\).

1. Write down the given data\
First term $a = 10$, number of terms $n = 14$, sum $S_{14}=1050$.

2. Use the sum formula of an AP\
$$S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr]$$
Substituting the known values:
$$1050 = \frac{14}{2}\bigl[2\times10 + (14-1)d\bigr]$$
$$1050 = 7\bigl[20 + 13d\bigr]$$

3. Solve for the common difference $d$\
Divide both sides by 7:
$$\frac{1050}{7} = 20 + 13d \quad\Rightarrow\quad 150 = 20 + 13d$$
$$13d = 150 - 20 = 130$$
$$d = \frac{130}{13} = 10$$

4. Find the 20th term using the nth‑term formula\
$$a_n = a + (n-1)d$$
For $n = 20$:
$$a_{20} = 10 + (20-1)\times10 = 10 + 19\times10 = 10 + 190 = 200$$

5. Answer\
The 20th term of the given AP is $\boxed{200}$.

1. Identify the first term \(a\) and common difference \(d\):
\[ a = 24, \quad d = 21-24 = -3. \]

2. Write the sum formula for \(n\) terms and set it equal to 78:
\[ S_n = \frac{n}{2}[2a + (n-1)d] = 78. \]
Substituting \(a\) and \(d\):
\[ \frac{n}{2}[2\times24 + (n-1)(-3)] = 78. \]

3. Simplify the expression inside the brackets:
\[ 2\times24 = 48, \quad (n-1)(-3) = -3n + 3. \]
Hence,
\[ \frac{n}{2}[48 - 3n + 3] = \frac{n}{2}[51 - 3n] = 78. \]

4. Multiply both sides by 2 to clear the denominator:
\[ n(51 - 3n) = 156. \]

5. Expand and bring all terms to one side:
\[ 51n - 3n^2 = 156 \]
\[ 3n^2 - 51n + 156 = 0. \]

6. Divide the quadratic by 3 for simplicity:
\[ n^2 - 17n + 52 = 0. \]

7. Factorise the quadratic:
\[ (n - 13)(n - 4) = 0. \]
Hence, \(n = 13\) or \(n = 4\).

8. Verify both values:
- For \(n = 4\):
\[ S_4 = \frac{4}{2}(24 + 15) = 2 \times 39 = 78. \]
- For \(n = 13\):
Last term \(a_{13} = 24 + 12(-3) = -12\).
\[ S_{13} = \frac{13}{2}(24 + (-12)) = \frac{13}{2} \times 12 = 13 \times 6 = 78. \]
Both satisfy the condition.

9. Therefore, the required number of terms can be either 4 or 13.

Step 1: Identify the series as an AP with first term $a=1$ and common difference $d=1$.
Step 2: For the first $n$ positive integers, the $n^{th}$ term (last term) is $l = a + (n-1)d = 1 + (n-1)\times1 = n$.
Step 3: Use the sum formula $S_n = \frac{n}{2}(a + l)$.
Step 4: Substitute $a=1$ and $l=n$:
$$S_n = \frac{n}{2}(1 + n) = \frac{n(n+1)}{2}.$$
Step 5: (i) For $n = 1000$,
$$S_{1000} = \frac{1000\times1001}{2} = 500\times1001 = 500500.$$
Step 6: (ii) For a general $n$, the sum is $S_n = \frac{n(n+1)}{2}$.
Thus the required sums are obtained.

1. Identify the AP parameters\
The nth term is \(a_n = 3 + 2n\).\
For \(n=1\): \(a_1 = 3 + 2\times1 = 5\). Hence the first term \(a = 5\).\
The common difference \(d\) is obtained from \(a_{n+1} - a_n\):\
\[d = (3+2(n+1)) - (3+2n) = 2\].\
So, \(d = 2\).\

2. Find the 24th (last) term\
\[l = a + (n-1)d = 5 + (24-1)\times2 = 5 + 46 = 51\].\

3. Apply the sum formula for an AP\
Using \(S_n = \frac{n}{2}(a + l)\):\
\[S_{24} = \frac{24}{2}(5 + 51) = 12 \times 56 = 672\].\
Alternatively, using \(S_n = \frac{n}{2}[2a + (n-1)d]\):\
\[S_{24} = \frac{24}{2}[2\times5 + (24-1)\times2] = 12[10 + 46] = 12 \times 56 = 672\].\

4. Result\
The sum of the first 24 terms is \(\boxed{672}\).

1. Let $a$ be the production in the 1st year (first term) and $d$ be the common increase each year.\
2. Using the term formula for the 3rd and 7th years:
$$a_3 = a + 2d = 600 \quad\text{(i)}$$
$$a_7 = a + 6d = 700 \quad\text{(ii)}$$
3. Subtract (i) from (ii):
$$ (a+6d) - (a+2d) = 700 - 600 \Rightarrow 4d = 100 \Rightarrow d = 25.$$\
4. Substitute $d = 25$ into (i) to find $a$:
$$a + 2(25) = 600 \Rightarrow a = 600 - 50 = 550.$$\
5. (i) Production in the 1st year: $a = \boxed{550}$ sets.\
6. (ii) Production in the 10th year:
$$a_{10} = a + 9d = 550 + 9\times25 = 550 + 225 = \boxed{775}\text{ sets}.$$\
7. (iii) Total production in the first 7 years:
Using $S_7 = \frac{7}{2}(a_1 + a_7)$:
$$S_7 = \frac{7}{2}(550 + 700) = \frac{7}{2}\times1250 = 7\times625 = \boxed{4375}\text{ sets}.$$\
8. All required quantities are obtained.