EXERCISE 5.2

### Row (i)
- Given: \(a = 7\), \(d = 3\), \(n = 8\).
- Use \(a_n = a + (n-1)d\):
\[a_8 = 7 + (8-1)\times 3 = 7 + 7\times 3 = 7 + 21 = 28.\]
- Hence \(a_n = 28\).

### Row (ii)
- Given: \(d = -18\), \(n = 10\), \(a_n = 0\). (The first entry \(a\) is missing.)
- Apply the formula:
\[0 = a + (10-1)(-18) = a - 9\times 18 = a - 162.\]
- Solve for \(a\): \(a = 162\).

### Row (iii)
- Given: \(d = -3\), \(n = 18\), \(a_n = -5\). (The first entry \(a\) is missing.)
- Use the formula:
\[-5 = a + (18-1)(-3) = a - 51.\]
- Hence \(a = 46\).

### Row (iv)
- Given: \(a = -18.9\), \(d = 2.5\), \(a_n = 3.6\). (The term \(n\) is missing.)
- Substitute:
\[3.6 = -18.9 + (n-1)\times 2.5\]
\[ (n-1)\times 2.5 = 3.6 + 18.9 = 22.5 \]
\[ n-1 = \frac{22.5}{2.5} = 9 \]
\[ n = 10.\]
- Therefore \(n = 10\).

### Row (v)
- Given: \(a = 3.5\), \(n = 105\), \(a_n = 62\). (The common difference \(d\) is missing.)
- Apply the formula:
\[62 = 3.5 + (105-1)d = 3.5 + 104d.\]
\[104d = 62 - 3.5 = 58.5\]
\[d = \frac{58.5}{104} = 0.5625 = \frac{9}{16}.\]
- Hence \(d = 0.5625\) (or \(\frac{9}{16}\)).

Filled table
| a | d | n | a_n |
|---|---|---|-----|
| 7 | 3 | 8 | 28 |
| 162 | -18 | 10 | 0 |
| 46 | -3 | 18 | -5 |
| -18.9 | 2.5 | 10 | 3.6 |
| 3.5 | 0.5625 | 105 | 62 |

(i)
1. Identify the first term \(a = 10\).
2. Find the common difference \(d = 7-10 = -3\).
3. Apply the nth‑term formula for \(n = 30\):
$$a_{30} = 10 + (30-1)(-3) = 10 + 29(-3) = 10 - 87 = -77.$$
4. Hence the 30th term is \(-77\). The correct choice is (C) –77.

(ii)
1. The given AP is \(-3,\; -\frac{1}{2},\; 2,\; \dots\). Thus the first term \(a = -3\).
2. Compute the common difference:
$$d = \left(-\frac{1}{2}\right) - (-3) = \frac{5}{2}.$$
(The same difference is confirmed by \(2 - (-\frac{1}{2}) = \frac{5}{2}\).)
3. Use the nth‑term formula for \(n = 11\):
$$a_{11} = -3 + (11-1)\left(\frac{5}{2}\right) = -3 + 10\times\frac{5}{2} = -3 + 25 = 22.$$
4. Hence the 11th term is \(22\). The correct choice is (B) 22.

1. Identify the known terms and the total number of terms (\(n\)).\
2. Compute the common difference \(d\) using\
$$d = \frac{\text{last term} - \text{first term}}{n-1}.$$\
3. Use the formula\
$$a_k = a_1 + (k-1)d$$\
to find each missing term.\

(i) 2, __ , 26\
- Here \(n = 3\), \(a_1 = 2\), \(a_3 = 26\).\
- \(d = \dfrac{26-2}{2} = 12\).\
- Missing term \(a_2 = a_1 + d = 2 + 12 = 14\).\

(ii) __ , 13, __ , 3\
- \(n = 4\), \(a_2 = 13\), \(a_4 = 3\).\
- Let \(d\) be the common difference.\
- From \(a_4 = a_1 + 3d\) and \(a_2 = a_1 + d\):\
\( (a_1+3d) - (a_1+d) = 3-13 \Rightarrow 2d = -10 \Rightarrow d = -5\).\
- \(a_1 = a_2 - d = 13 - (-5) = 18\).\
- \(a_3 = a_2 + d = 13 + (-5) = 8\).\

(iii) 5, __ , __ , 192\
- \(n = 4\), \(a_1 = 5\), \(a_4 = 192\).\
- \(d = \dfrac{192-5}{3} = \dfrac{187}{3}\).\
- \(a_2 = a_1 + d = 5 + \dfrac{187}{3} = \dfrac{202}{3}\).\
- \(a_3 = a_2 + d = \dfrac{202}{3} + \dfrac{187}{3} = \dfrac{389}{3}\).\

(iv) –4, __ , __ , __ , __ , 6\
- \(n = 6\), \(a_1 = -4\), \(a_6 = 6\).\
- \(d = \dfrac{6-(-4)}{5} = \dfrac{10}{5} = 2\).\
- Successive terms: \(a_2 = -4+2 = -2\), \(a_3 = 0\), \(a_4 = 2\), \(a_5 = 4\).\

(v) __ , 38, __ , __ , __ , –22\
- \(n = 6\), \(a_2 = 38\), \(a_6 = -22\).\
- Let \(d\) be the common difference.\
- From \(a_6 = a_1 + 5d\) and \(a_2 = a_1 + d\):\
\((a_1+5d) - (a_1+d) = -22 - 38 \Rightarrow 4d = -60 \Rightarrow d = -15\).\
- \(a_1 = a_2 - d = 38 - (-15) = 53\).\
- \(a_3 = a_2 + d = 38 - 15 = 23\).\
- \(a_4 = a_3 + d = 23 - 15 = 8\).\
- \(a_5 = a_4 + d = 8 - 15 = -7\).

1. Identify the first term and common difference of the given AP.
- First term \(a = 3\).
- Common difference \(d = 8-3 = 5\).
2. Write the nth term formula for the AP:
$$a_n = a + (n-1)d$$
3. Substitute \(a = 3\), \(d = 5\) and \(a_n = 78\) into the formula:
$$78 = 3 + (n-1)\times 5$$
4. Solve for \(n\):
\[78 - 3 = (n-1)\times 5\]
\[75 = 5(n-1)\]
\[n-1 = \frac{75}{5} = 15\]
\[n = 15 + 1 = 16\]
5. Hence, the number 78 occurs at the 16th position in the given AP.

### (i) \(7,\;13,\;19,\;\dotsc,\;205\)
1. Identify the first term \(a\) and the common difference \(d\).
- \(a = 7\)
- \(d = 13 - 7 = 6\)
2. Use the formula for the number of terms:
\[ n = \frac{l - a}{d} + 1 \]
where \(l = 205\).
3. Substitute the values:
\[ n = \frac{205 - 7}{6} + 1 = \frac{198}{6} + 1 = 33 + 1 = 34 \]
4. Hence, the AP contains 34 terms.

### (ii) \(18,\;\frac{15}{2},\;13,\;\dotsc,\;-47\)
1. First term \(a = 18\).
2. The common difference \(d\) can be obtained from the first and third terms (since the second term given is ambiguous, we use the first and third terms which are definitely part of the AP):
\[ d = \frac{13 - 18}{2} = -\frac{5}{2} \]
3. Apply the formula for the number of terms with \(l = -47\):
\[ n = \frac{-47 - 18}{-\frac{5}{2}} + 1 \]
4. Simplify the numerator:
\[ -47 - 18 = -65 \]
5. Divide by \(-\frac{5}{2}\):
\[ \frac{-65}{-\frac{5}{2}} = -65 \times \left(-\frac{2}{5}\right) = \frac{130}{5} = 26 \]
6. Add 1:
\[ n = 26 + 1 = 27 \]
7. Therefore, the AP contains 27 terms.

Note: The second AP in the question has a typographical inconsistency in the second term. Using the first and third terms (which are clearly part of the progression) yields a consistent common difference and the above result.

1. Identify the first term and common difference of the given AP.
\[ a = 11, \quad d = 8-11 = -3 \]
2. Write the general (nth) term of the AP.
\[ a_n = a + (n-1)d = 11 + (n-1)(-3) = 11 - 3(n-1) = 14 - 3n \]
3. Set $a_n$ equal to the required term $-150$ and solve for $n$.
\[ -150 = 14 - 3n \]
\[ -150 - 14 = -3n \]
\[ -164 = -3n \]
\[ n = \frac{164}{3} \]
4. Since $n = \frac{164}{3} \approx 54.67$ is not a positive integer, the required term cannot be obtained from the AP.
5. Conclude that $-150$ is not a term of the given AP.

1. Let $a$ be the first term and $d$ the common difference of the AP.\
2. Using the $n^{\text{th}}$ term formula $a_n = a + (n-1)d$:
$$\begin{aligned}
a_{11} &= a + 10d = 38 \quad\text{(i)}\\
a_{16} &= a + 15d = 73 \quad\text{(ii)}\end{aligned}$$
3. Subtract equation (i) from equation (ii) to eliminate $a$:
$$\begin{aligned}(a + 15d) - (a + 10d) &= 73 - 38 \\
5d &= 35 \\
\Rightarrow \; d &= \frac{35}{5} = 7\end{aligned}$$
4. Substitute $d = 7$ back into equation (i) to find $a$:
$$\begin{aligned}a + 10(7) &= 38 \\
a + 70 &= 38 \\
\Rightarrow \; a &= 38 - 70 = -32\end{aligned}$$
5. Now find the 31st term $a_{31}$:
$$\begin{aligned}a_{31} &= a + (31-1)d \\
&= -32 + 30 \times 7 \\
&= -32 + 210 \\
&= 178\end{aligned}$$
6. Hence, the 31st term of the given AP is 178.

1. Let the first term be $a$ and the common difference be $d$.
2. Given:
- 3rd term $a_3 = 12$ \[ $a_3 = a + 2d = 12$ \]
- Last term (50th term) $a_{50} = 106$ \[ $a_{50} = a + 49d = 106$ \]
3. Express $a$ from the first condition:
$$a = 12 - 2d$$
4. Substitute $a$ into the second condition:
$$ (12 - 2d) + 49d = 106$$
$$ 12 + 47d = 106$$
$$ 47d = 94$$
$$ d = \frac{94}{47} = 2$$
5. Find the first term $a$:
$$ a = 12 - 2d = 12 - 2\times 2 = 12 - 4 = 8$$
6. Find the 29th term using $a_{29} = a + 28d$:
$$ a_{29} = 8 + 28\times 2 = 8 + 56 = 64$$
7. Result: The 29th term of the AP is $64$.

1. Let the first term be $a$ and the common difference be $d$.
2. Using the formula $T_n = a + (n-1)d$:
- For the 3rd term: $T_3 = a + 2d = 4$ … (i)
- For the 9th term: $T_9 = a + 8d = -8$ … (ii)
3. Subtract (i) from (ii):
$$ (a+8d) - (a+2d) = -8 - 4 \[4pt] 6d = -12 \[4pt] d = -2 $$
4. Substitute $d = -2$ into (i):
$$ a + 2(-2) = 4 \[4pt] a - 4 = 4 \[4pt] a = 8 $$
5. The AP is therefore $a = 8$, $d = -2$.
6. To find the term which is zero, set $T_n = 0$:
$$ a + (n-1)d = 0 \[4pt] 8 + (n-1)(-2) = 0 \[4pt] 8 - 2(n-1) = 0 \[4pt] 2(n-1) = 8 \[4pt] n-1 = 4 \[4pt] n = 5 $$
7. Hence the 5th term of the AP is zero.

1. Let $a$ be the first term and $d$ be the common difference of the AP.
2. Write the expressions for the 17th and 10th terms using the $n^{\text{th}}$ term formula:
$$a_{17} = a + (17-1)d = a + 16d,$$
$$a_{10} = a + (10-1)d = a + 9d.$$
3. According to the question, the 17th term exceeds the 10th term by 7:
$$a_{17} - a_{10} = 7.$$
4. Substitute the expressions from step 2:
$$(a + 16d) - (a + 9d) = 7.$$
5. Simplify the left‑hand side:
$$a + 16d - a - 9d = 7 \[2mm] 7d = 7.$$
6. Solve for $d$:
$$d = \frac{7}{7} = 1.$$
7. Hence, the common difference of the AP is $\boxed{1}$.

1. Identify the first term and common difference:
\[ a = 3, \quad d = 15-3 = 12. \]
2. Write the general term of the AP:
\[ T_n = a + (n-1)d = 3 + (n-1)\times12 = 12n - 9. \]
3. Find the 54th term using the formula:
\[ T_{54} = 12\times54 - 9 = 648 - 9 = 639. \]
4. According to the problem, the required term $T_n$ satisfies:
\[ T_n = T_{54} + 132. \]
Substitute $T_{54}=639$:
\[ T_n = 639 + 132 = 771. \]
5. Set the general expression equal to 771 and solve for $n$:
\[ 12n - 9 = 771 \]
\[ 12n = 771 + 9 = 780 \]
\[ n = \frac{780}{12} = 65. \]
6. Hence the 65th term of the given AP is 132 more than its 54th term.

Answer: The 65th term.

1. Let the first AP be \(a_1, a_1+d, a_1+2d,\dots\) and the second AP be \(a_2, a_2+d, a_2+2d,\dots\).\
2. General term of an AP: \(T_n = a + (n-1)d\).\
3. 100th term of the first AP: \(T_{100}^{(1)} = a_1 + 99d\).\
100th term of the second AP: \(T_{100}^{(2)} = a_2 + 99d\).\
4. Given \(T_{100}^{(1)} - T_{100}^{(2)} = 100\):\
\[ (a_1 + 99d) - (a_2 + 99d) = a_1 - a_2 = 100. \]
Hence, \(a_1 - a_2 = 100\).\
5. 1000th term of the first AP: \(T_{1000}^{(1)} = a_1 + 999d\).\
1000th term of the second AP: \(T_{1000}^{(2)} = a_2 + 999d\).\
6. Difference between the 1000th terms:\
\[ T_{1000}^{(1)} - T_{1000}^{(2)} = (a_1 + 999d) - (a_2 + 999d) = a_1 - a_2. \]
Using the result from step 4, \(a_1 - a_2 = 100\).\
7. Therefore, the difference between the 1000th terms is also \(100\).

1. Identify the smallest three‑digit multiple of 7\
\(100 \div 7 = 14\) remainder 2, so the next multiple is \(7\times 15 = 105\).\
Hence, the first term \(a = 105\).

2. Identify the largest three‑digit multiple of 7\
\(999 \div 7 = 142\) remainder 5, so the greatest multiple not exceeding 999 is \(7\times 142 = 994\).\
Hence, the last term \(l = 994\).

3. Common difference\
Since we are dealing with consecutive multiples of 7, the common difference \(d = 7\).

4. Number of terms in the AP\
Use the formula for the number of terms of an AP:
$$n = \frac{l - a}{d} + 1$$
Substituting the values:
$$n = \frac{994 - 105}{7} + 1 = \frac{889}{7} + 1 = 127 + 1 = 128$$

5. Conclusion\
Therefore, there are \(128\) three‑digit numbers that are divisible by 7.

1. Identify the AP formed by multiples of 4: \(4, 8, 12, 16, \dots\) with common difference \(d = 4\).
2. Find the smallest multiple of 4 greater than 10. Since \(4 \times 2 = 8 < 10\) and \(4 \times 3 = 12 > 10\), the first term \(a = 12\).
3. Find the largest multiple of 4 less than 250. Since \(4 \times 62 = 248 < 250\) and \(4 \times 63 = 252 > 250\), the last term \(l = 248\).
4. Use the formula for the number of terms in an AP:
$$ n = \frac{l - a}{d} + 1 $$
Substituting the values:
$$ n = \frac{248 - 12}{4} + 1 = \frac{236}{4} + 1 = 59 + 1 = 60 $$
5. Hence, there are 60 multiples of 4 lying between 10 and 250.

1. Identify the first term and common difference of each AP.
- First AP: $a_1 = 63$, $d_1 = 65-63 = 2$.
- Second AP: $a_2 = 3$, $d_2 = 10-3 = 7$.

2. Write the expression for the nth term of each AP using $a_n = a + (n-1)d$.
- For the first AP: $$a_n = 63 + (n-1) \times 2 = 2n + 61.$$
- For the second AP: $$b_n = 3 + (n-1) \times 7 = 7n - 4.$$

3. Set the two nth terms equal because we are asked when they are the same.
$$2n + 61 = 7n - 4.$$

4. Solve the linear equation for $n$.
\[\begin{aligned}
2n + 61 &= 7n - 4 \\
61 + 4 &= 7n - 2n \\
65 &= 5n \\
n &= \frac{65}{5} = 13.
\end{aligned}\]

5. Verify by substituting $n = 13$ back into both nth‑term formulas.
- First AP: $a_{13} = 2(13) + 61 = 87$.
- Second AP: $b_{13} = 7(13) - 4 = 87$.
Both give the same value, confirming the solution.

Let the first term of the A.P. be $a$ and the common difference be $d$.

1. Third term condition
\[ T_3 = a + 2d = 16 \]
This gives the first equation:
\[ a + 2d = 16 \] (Equation 1)

2. Difference between 7th and 5th terms
\[ T_7 - T_5 = (a + 6d) - (a + 4d) = 2d = 12 \]
Hence,
\[ 2d = 12 \]
\[ d = 6 \] (Equation 2)

3. Find the first term
Substitute $d = 6$ into Equation 1:
\[ a + 2(6) = 16 \]
\[ a + 12 = 16 \]
\[ a = 4 \]

4. Write the required A.P.
The A.P. is:
\[ \{a, a+d, a+2d, a+3d, \dots\} = \{4, 4+6, 4+2\times6, 4+3\times6, \dots\} = \{4, 10, 16, 22, 28, 34, 40, \dots\} \]

5. Verification
- Third term $= 4 + 2\times6 = 16$ (as given).
- 7th term $= 4 + 6\times6 = 40$, 5th term $= 4 + 4\times6 = 28$.
- Difference $= 40 - 28 = 12$, satisfying the second condition.

Thus the required A.P. has first term $a = 4$ and common difference $d = 6$.

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

2. Let the total number of terms be \(n\). The last term given is \(a_n = 253\). Use the nth‑term formula:
\[ a_n = a + (n-1)d \]
\[ 253 = 3 + (n-1)\times5 \]
\[ (n-1)\times5 = 250 \]
\[ n-1 = 50 \]
\[ n = 51. \]
Hence the AP contains 51 terms.

3. "20th term from the last term" means we count the last term as the 1st term from the end. Therefore the required term is the \( (n-20+1)^{\text{th}} \) term:
\[ \text{Position} = 51 - (20-1) = 32. \]

4. Find the 32nd term using the nth‑term formula:
\[ a_{32} = a + (32-1)d \]
\[ a_{32} = 3 + 31\times5 \]
\[ a_{32} = 3 + 155 = 158. \]

5. Hence the 20th term from the last term of the given AP is 158.

1. Write the required terms
\[
t_4 = a + 3d, \quad t_8 = a + 7d,\quad t_6 = a + 5d, \quad t_{10}= a + 9d.
\]

2. Use the given information
\[
t_4 + t_8 = (a+3d) + (a+7d) = 2a + 10d = 24. \tag{1}
\]
From (1) we obtain
\[
a + 5d = 12. \tag{2}
\]
(Equation (2) is obtained by dividing (1) by 2.)

3. Express the required sum
\[
t_6 + t_{10} = (a+5d) + (a+9d) = 2a + 14d. \tag{3}
\]

4. Replace $2a$ using (1)
From (1), $2a = 24 - 10d$. Substituting in (3):
\[
t_6 + t_{10} = (24 - 10d) + 14d = 24 + 4d. \tag{4}
\]

5. Conclusion
The sum of the 6th and 10th terms cannot be fixed to a single numeric value unless the common difference $d$ is known. It is given by the expression
\[
\boxed{\;t_6 + t_{10} = 24 + 4d\;}.\]
If additional information (for example, the value of $d$) were supplied, the numerical value could be obtained.

6. Remark (useful for exam)
In many CBSE problems the extra condition is provided (e.g., the sum of the 6th and 10th terms is 28). In that case one would set $24+4d = 28$ to find $d = 1$, and then $a = 12 - 5d = 7$.
However, with the data given in the present statement, the answer remains the expression $24+4d$.

1. Identify the AP:
- First term (salary in 1995): $a = 5000$.
- Common difference (annual increment): $d = 200$.

2. Let $a_n$ be the salary in the $n^{\text{th}}$ year after 1995. The $n^{\text{th}}$ term of an AP is given by:
$$a_n = a + (n-1)d$$

3. Set $a_n = 7000$ and solve for $n$:
$$7000 = 5000 + (n-1)\times 200$$
$$7000 - 5000 = (n-1)\times 200$$
$$2000 = 200\,(n-1)$$
$$n-1 = \frac{2000}{200} = 10$$
$$n = 10 + 1 = 11$$

4. The first term corresponds to the year 1995, so the $n^{\text{th}}$ term corresponds to the year:
$$\text{Year} = 1995 + (n-1) = 1995 + 10 = 2005$$

5. Hence, Subba Rao's salary becomes `7000` in the year 2005.

1. Identify the given data:
- First week saving (first term) \(a = \text{` }5\).
- Weekly increase (common difference) \(d = \text{` }1.75\).
- Saving in the \(n\)th week (nth term) \(a_n = \text{` }20.75\).

2. Write the nth term formula for an AP:
$$a_n = a + (n-1)d$$

3. Substitute the known values:
$$20.75 = 5 + (n-1)\times 1.75$$

4. Isolate \((n-1)\):
$$20.75 - 5 = (n-1)\times 1.75$$
$$15.75 = (n-1)\times 1.75$$

5. Divide both sides by \(1.75\):
$$n-1 = \frac{15.75}{1.75}$$
Since \(1.75 \times 9 = 15.75\), we get
$$n-1 = 9$$

6. Solve for \(n\):
$$n = 9 + 1 = 10$$

Therefore, Ramkali reaches a weekly saving of `20.75 in the 10th week.

1. Let the first term be \(a\) and the common difference be \(d\).\
2. Sum of first three terms (\(S_3\)) is given as 12. Using the sum formula:
$$S_3 = \frac{3}{2}[2a+(3-1)d] = \frac{3}{2}(2a+2d) = 3(a+d) = 12$$
Hence, \(a + d = 4\).\
3. Sum of first six terms (\(S_6\)) is given as 42. Using the sum formula:
$$S_6 = \frac{6}{2}[2a+(6-1)d] = 3(2a+5d) = 42$$
Hence, \(2a + 5d = 14\).\
4. Solve the two linear equations:
\[\begin{cases} a + d = 4 \ 2a + 5d = 14 \end{cases}\]
Multiply the first equation by 2: \(2a + 2d = 8\).
Subtract from the second equation:
\( (2a+5d) - (2a+2d) = 14 - 8 \)\
\(3d = 6 \Rightarrow d = 2\).
Substitute \(d = 2\) into \(a + d = 4\): \(a + 2 = 4 \Rightarrow a = 2\).
5. The first three terms of the AP are:
\[a = 2,\quad a+d = 2+2 = 4,\quad a+2d = 2+2\times2 = 6\].