**Q.1: The first and the last term of an AP are 5 and 45 respectively. If the sum of all its terms is 400, then find the common difference and the number of terms.**

**Sol:**

Suppose there are n terms in the AP.

Here, a = 5, l = 45 and \(S_{n} = 400\)

\(S_{n} = 400 = \frac{n}{2} (5 + 45)\)

\(\Rightarrow \frac{n}{2} (50) = 400\)

\(\Rightarrow n = 16\)

Thus, there are 16 terms in the AP.

Let d be the common difference of the AP

\(Therefore, a_{16} = 45\)

\(\Rightarrow 5 + (16 – 1) \times d = 45\)

\(\Rightarrow 15 d = 40\)

\(\Rightarrow d = \frac{8}{3}\)

Hence the common difference of the AP is \(\frac{8}{3}\)

**Q.2: In an AP, the first term is 22, nth term is -11 and sum of first n terms is 66. Find n and hence find the common difference. **

**Sol:**

Here a = 22, \(T_{n} = -11 \;\; and \;\; S_{n} = 66\)

Let d be the common difference of the given AP.

Then \(T_{n} = -11\)

\(\Rightarrow a + (n – 1)d = 22 + (n – 1)d = -11\)

\(\Rightarrow (n – 1)d = – 33\)

The sum of n terms of an AP is given by

\(S_{n} = \frac{n}{2} (2a + (n – 1) d) = 66\)

\(\Rightarrow \frac{n}{2} (2 \times 22 + (-33)) = \left ( \frac{n}{2} \times 11\right ) = 66\)

\(\Rightarrow n = 12\)

Putting the value of n in (i), we get:

\(11d = -33\)

\(\Rightarrow d = -3\)

Thus, n = 12 and d = -3

**Q.3: The 12 ^{th} term of an AP is -13 and the sum of its first four terms is 24. Find the sum of its first 10 terms.**

**Sol:**

Let a be the first term and d be the common difference of the AP.

Then,

\(a_{12} = – 13\)

\(\Rightarrow a + 11 d = – 13\)

Also,

\(S_{4} = 24\)

\(\Rightarrow \frac{4}{2} (2a + 3d) = 24\)

\(\Rightarrow (2a + 3d) = 12\)

Solving (i) and (ii) we get:

\(2(- 13 – 11d) + 3d = 12\)

\(\Rightarrow – 26 – 22d + 3d = 12\)

\(\Rightarrow – 19 d = 12 + 26 = 38\)

\(\Rightarrow d = -2\)

Putting d = -2 in (i) we get:

\(\Rightarrow a + 11 ( – 2) = – 13\)

\(\Rightarrow a = -13 + 22 = 9\)

Therefore, Sum of its first 10 terms, \(S_{10}\)

\(= \frac{10}{2} \left [ 2 \times 9 + (10 – 1) \times (-2) \right ]\)

\(= 5 \times (18 – 18)\)

\(= 5 \times 0\)

= 0

Hence, the required sum is 0.

**Q.4: The sum of the first 7 terms of an AP is 182. If its 4 ^{th} and 17^{th} terms are in the ratio 1 : 5, find the AP.**

**Sol:**

Let a be the first term and d be the common difference of the AP.

\(Therefore, S_{7} = 182\)

\(\Rightarrow \frac{7}{2} (2a + 6d) = 182\)

\(\Rightarrow a + 3d = 26\)

Also,

\(a_{4} : a_{17} = 1 : 5\)

\(\Rightarrow \frac{a + 3d}{a + 16d} = \frac{1}{5}\)

\(\Rightarrow 5a + 15 d = a + 16d\)

\(\Rightarrow d = 4a\)

Solving (i) and (ii), we get

\(a + 3 \times 4 a = 26\)

\(\Rightarrow 13 a = 26\)

\(\Rightarrow a = 2 \)

Putting a = 2 in (ii), we get

\(d = 4 \times 2 = 8\)

Hence, the required AP is 2, 10 , 18, 26, ……….

**Q.5: The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first term and the common difference of the AP.**

**Sol:**

Here a= 4 , d = 7 and l = 81

Let the nth term be 81

Then \(T_{n} = 81\)

\(\Rightarrow a + (n – 1)d = 4 + (n – 1)7 = 81\)

\(\Rightarrow (n – 1)7 = 77\)

\(\Rightarrow (n – 1) = 11\)

\(\Rightarrow n = 12\)

Thus, there are 12 terms in the AP.

The sum of n terms of an AP is given by

\(S_{n} = \frac{n}{2} (a + l)\)

\(S_{12} = \frac{12}{2} (4 + 81) = 6 \times 85 = 510\)

Thus, the required sum is 510.

**Q.6: The sum of the first 7 terms of an AP is 49 and the sum of its first 17 terms is 289. Find the sum of its first n terms. **

**Sol:**

Let a be the first term and d be the common difference of the given AP.

Then we have:

\(S_{n} = \frac{n}{2} \left [ 2a + (n – 1)d \right ]\)

\(S_{7} = \frac{7}{2} \left [ 2a + 6d \right ] = 7 \left [ a + 3d \right ]\)

\(S_{17} = \frac{17}{2} \left [ 2a + 6d \right ] = 17 \left [ a + 3d \right ]\)

However, \(S_{7} = 49 \;\; and \;\;S_{17} = 289\)

However, \(S_{7} = 49 \;\; and \;\;S_{17} = 289\)

Now, \(7 \left [ a + 3d \right ] = 49\)

\(\Rightarrow a + 3d = 7\)

Also, \(17 \left [ a + 8d \right ] = 289\)

\(\Rightarrow a + 8 d = 17\)

Subtracting (i) from (ii), we get:

5d = 10

\(\Rightarrow d = 2\)

Putting d = 2 in (i), we get:

a + 6 = 7

\(\Rightarrow a = 1\)

Thus, a = 1 and d = 2

Therefore, Sum of n terms of AP = \(\frac{n}{2} \left [ 2 \times 1 + (n -1) \times 2 \right ] = n \left [ 1 + (n – 1) \right ] = n^{2}\)

**Q.7: Two APs have the same common difference. If the first terms of these APs are 3 and 8 respectively, find the difference between the sums of their first 50 terms.**

**Sol:**

Let \(a_{1} \;\; and \;\; a_{2}\)

Here, \(a_{1} = 8\)

Suppose d be the common difference of two AP.

Let \(S_{50} \;\; and \;\; {S}’_{50}\)

\(Therefore, S_{50} – {S}’_{50} = \frac{50}{2} \left [ 2 a_{1} + (50 – 1) d\right ] – \frac{50}{2} \left [ 2 a_{2} + (50 – 1) d\right ]\)

\(= 25 \left [ 2 \times 8 + 49 d\right ] – 25 \left [ 2 \times 3 + 49 d\right ]\)

\(= 25 \times (16 – 6)\)

= 250

Hence, the required difference between the two sums is 250.

**Q.8: The sum of first 10 terms of an AP is -150 and the sum of its next 10 terms is -550. Find the AP.**

**Sol:**

Let a be the first term and d be the common difference of the AP.

Then,

\(S_{10} = -150\)

\(\Rightarrow \frac{10}{2} (2a + 9d) = -150\)

\(\Rightarrow 5 (2a + 9d) = -150\)

\(\Rightarrow (2a + 9d) = -30\)

It is given that the sum of its next 10 terms is – 550

Now,

\(S_{20}\)

\(Therefore, S_{20} = -700\)

\(\Rightarrow \frac{20}{2} (2a + 19d) = – 700\)

\(\Rightarrow 10 (2a + 19d) = – 700\)

\(\Rightarrow 2a + 19d = – 70\)

Subtracting (i) from (ii), we get

\(2a + 19d – (2a + 9d) = – 70 – (-30)\)

\(\Rightarrow 10 d = -40\)

\(\Rightarrow d = -4\)

Putting d = -4 in (i), we get

\(2a + 9 \times (- 4) = – 30\)

\(\Rightarrow 2a = – 30 + 36 = 6\)

\(\Rightarrow a = 3\)

Hence, the required AP is 3, -1, -5, -9, ……

**Q.9: The 13 ^{th} term of an AP is 5 times its 3^{rd} term. If its 5^{th} term is 16, find the sum of its first 10 terms.**

**Sol:**

Let a be the first term and d be the common difference of the AP. Then,

\(a_{13} = 4 \times a_{3}\)

\(\Rightarrow a + 12 d = 4 (a + 2d)\)

\(\Rightarrow a + 12 d = 4 a + 8d\)

\(\Rightarrow 3a = 4d\)

Also,

\(a_{5} = 16\)

\(\Rightarrow a + 4d = 16\)

Solving (i) and (ii), we get

\(a + 3a = 12\)

\(4a = 12\)

\(a = 3\)

Putting a = 4 in (i) we get:

\(4d = 3 \times 4 = 12\)

\(\Rightarrow d = 3\)

Using the formula,

\(S_{n} = \frac{n}{2} \left [ 2a + (n – 1)d \right ]\)

\(S_{10} = \frac{10}{2} \left [ 2 \times 4 + (10 – 1) \times 3 \right ]\)

\(= 5 \left [ 8 + 27 \right ]\)

\(= 5 \left [35 \right ]\)

\(= 175\)

Hence the required sum is 175.

**Q.10: The 16 ^{th} term of an AP is 5 times its 3^{rd} term. If its 10^{th} term is 41, find the sum of its first 10 terms.**

**Sol:**

Let a be the first term and d be the common difference of the AP.

Then,

\(a_{16} = 5 \times a_{3}\)

\(\Rightarrow a + 15 d = 5 (a + 2d)\)

\(\Rightarrow a + 12 d = 5a + 10d\)

\(\Rightarrow 4a = 5d\)

Also,

\(a_{10} = 41\)

\(\Rightarrow a + 9d = 41\)

Solving (i) and (ii), we get

\(a + 9 \times \frac{4}{5}a = 41 \)

\(\Rightarrow \frac{5a + 36a}{5} = 41\)

\(\Rightarrow \frac{41a}{5} = 41\)

\( a = 5 \)

Putting a = 5 in (i) we get:

\( 5d = 4 \times 5 = 20 \)

\(\Rightarrow d = 4 \)

Using the formula,

\(S_{n} = \frac{n}{2} \left [ 2a + (n – 1)d \right ]\)

\(S_{15} = \frac{15}{2} \left [ 2 \times 5 + (15 – 1) \times 4 \right ]\)

\(= \frac{15}{2} \left [ 10 + 56 \right ]\)

\(= \frac{15}{2} \left [66 \right ]\)

\(= 495 \)

Hence the required sum is 495.

**Q.11: An AP 5, 12, 19, … has 50 terms. Find its last term. Hence, find the sum of its last 15 terms.**

**Sol:**

The given AP is 5, 12, 19, ………..

Here, a = 5, d = 12 – 5 = 7 and n = 50.

Since there are 50 terms in the AP, so the last term of the AP is \(a_{50}\)

l = \(a_{50}\)

= 5 + 343

= 348

Thus, the last term of the AP is 348.

Now,

Sum of the last 15 term of the AP

= \(S_{50} – S_{35} \)

\(= \frac{50}{2} \left [ 2 \times 5 + (50 – 1) \times 7 \right ] – \frac{35}{2} \left [ 2 \times 5 + (35 – 1) \times 7 \right ]\)

= \(= \frac{50}{2} \left [ 10 + 343 \right ] – \frac{35}{2} \left [ 10 + 238 \right ]\)

\(= (25 \times 353) – (35 \times 124)\)

\(= 4485\)

Hence, the required sum is 4485.

**Q.12: An AP 8, 10, 12, … has 60 terms. Find its last term. Hence, find the sum of its last 10 terms.**

**Sol:**

The given AP is 8, 10, 12, …………

Here, a = 8 , d = 10 – 8 = 2 and n = 60

Since there are 60 terms in the AP, so the last term of the AP is \(a_{50}\)

l = \(a_{50}\)

Thus, the last term of the AP is 126.

Now,

Sum of the last 10 terms of the AP

= \(S_{60} – S_{50} \)

\(= \frac{50}{2} \left [ 2 \times 5 + (50 – 1) \times 7 \right ] – \frac{35}{2} \left [ 2 \times 5 + (35 – 1) \times 7 \right ]\)

= \(= \frac{50}{2} \left [ 10 + 343 \right ] – \frac{35}{2} \left [ 10 + 238 \right ]\)

\(= (25 \times 353) – (35 \times 124)\)

\(= 4485\)

Hence, the required sum is 4485.

**Q.13: The sum of the 4 ^{th} and 8^{th} terms of AP is 24 and the sum of its 6^{th} and 10^{th} terms is 44. Find the sum of its first 10 terms. **

**Sol:**

Let a be the first term and d be the common difference of the AP.

\(Therefore, a_{4} + a_{8} = 24\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Also,

\(Therefore, a_{6} + a_{10} = 44\)

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Subtracting (i) from (ii), we get

(a + 7d) – (a + 5d) = 22 -12

\(\Rightarrow 2d = 10\)

\(\Rightarrow d = 5 \)

Putting d = 5 in (i), we get

a + 5\(\times\)

\(\Rightarrow\)

Using the formula, \(S_{n} = \frac{n}{2} \left [ 2a + (n – 1)d \right ]\)

\(= \frac{10}{2} \left [ 2 \times (-13) + (10 – 1) \times 5 \right ] \)

= \(= 5 \times (- 26 + 45 ) \)

\(= 5 \times 19 \)

\(= 95 \)

Hence, the required sum is 95.

**Q.14: The sum of first m terms of an AP is ( \(4m^{ 2 }\) – m ). If its nth term is 107, find the value of n. Also, find the 21 ^{st} term of this AP.**

**Sol:**

Let \(S_{m}\)

Then,

\(S_{m} = 4m^{2} – m\)

\(\Rightarrow S_{m-1} = 4(m – 1)^{2} – (m – 1)\)

\(= 4(m^{2} – 2m + 1) – (m – 1)\)

\(= 4m^{2} – 9m + 5\)

Suppose \(a_{m}\)

\(Therefore, a_{m} \)

\(= (4m^{2} – m) – (4m^{2} – 9m + 5 )\)

= 8m – 5 ……………….(i)

Now,

\(a_{n}\)

\(\Rightarrow 8n – 5 = 107\)

\(\Rightarrow 8n = 107 + 5 = 112 \)

\(\Rightarrow n = 14\)

Thus, the value of n is 14.

Putting m = 21 in (i), we get

\(a_{21} = 8 \times 21 – 5 = 168 – 5 = 163\)

Hence, the 21st term of the AP is 163.