Arithmetic Progression

Formula Used.

a_n = a + (n–1)d

If a = 3, d = 2

Then a₁ = a + (n–1)d = 3 + (1–1)2 = 3

a₂ = a + (n–1)d = 3 + (2–1)2 = 3 + 2×1 = 5

a₃ = a + (n–1)d = 3 + (3–1)2 = 3 + 2×2 = 7

a₄ = a + (n–1)d = 3 + (4–1)2 = 3 + 2×3 = 9

a_n = a + (n–1)d = 3 + (n–1)2 = 2n + 1

∴ The A.P is 3, 5, 7, 9……, 2n + 1

Formula Used.

a_n = a + (n–1)d

If a = –3, d = –2

Then a₁ = a + (n–1)d = –3 + (1–1)(–2) = –3

a₂ = a + (n–1)d = –3 + (2–1)(–2) = –3 + 1×(–2) = –5

a₃ = a + (n–1)d = –3 + (3–1)(–2) = –3 + 2×(–2) = –7

a₄ = a + (n–1)d = –3 + (4–1)(–2) = –3 + 3×(–2) = –9

a_n = a + (n–1)d = –3 + (n–1)(–2) = –1–2n

∴ The A.P is –3, –5, –7, –9……, –1–2n

Formula Used.

a_n = a + (n–1)d

If a = 100, d = –7

Then a₁ = a + (n–1)d = 100 + (1–1)(–7) = 100

a₂ = a + (n–1)d = 100 + (2–1)(–7) = 100 + (–7)×1 = 93

a₃ = a + (n–1)d = 100 + (3–1)(–7) = 100 + (–7)×2 = 86

a₄ = a + (n–1)d = 100 + (4–1)(–7) = 100 + (–7)×3 = 79

a_n = a + (n–1)d = 100 + (n–1)(–7) = 107 – 7n

∴ The A.P is 100, 93, 86, 79……107–7n

Formula Used.

a_n = a + (n–1)d

If a = –100, d = 7

Then a₁ = a + (n–1)d = –100 + (1–1)7 = –100

a₂ = a + (n–1)d = –100 + (2–1)7 = –100 + 7×1 = –93

a₃ = a + (n–1)d = –100 + (3–1)7 = –100 + 7×2 = –86

a₄ = a + (n–1)d = –100 + (4–1)7 = –100 + 7×3 = –79

a_n = a + (n–1)d = –100 + (n–1)7 = 7n–107

∴ The A.P is –100, –93, –86, –79……, 7n–107

Formula Used.

a_n = a + (n–1)d

If a = 1000, d = –100

Then;

a₁ = a + (n–1)d = 1000 + (1–1)(–100) = 1000

a₂ = a + (n–1)d = 1000 + (2–1)(–100) = 1000 + (–100)×1 = 900

a₃ = a + (n–1)d = 1000 + (3–1)(–100) = 1000 + (–100)×2 = 800

a₄ = a + (n–1)d = 1000 + (4–1)(–100) = 1000 + (–100)×3 = 700

a_n = a + (n–1)d = 1000 + (n–1)(–100) = 1100–100n

∴ The A.P is 1000, 900, 800, 700……, 1100–100n

Formula Used.

d_n = a_{n + 1} – a_n

In the above sequence,

a = 5;

d₁ = a₂–a₁ = (–5)–5 = –10

d₂ = a₃–a₂ = 5–(–5) = 10

d₃ = a₄–a₃ = (–5)–5 = –10

⇒ As in A.P the difference between the 2 terms is always constant

But the difference in sequence is not the same.

∴ The above sequence is not A.P

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 2;

d₁ = a₂–a₁ = 2–2 = 0

d₂ = a₃–a₂ = 2–2 = 0

d₃ = a₄–a₃ = 2–2 = 0

⇒ As in A.P the difference between the 2 terms is always constant

The difference in sequence comes to be 0.

∴ The above sequence is not an A.P

Formula Used.

d_n = a_{n + 1} – a_n

In the above sequence,

a = 1;

d₁ = a₂–a₁ = 11–1 = 10

d₂ = a₃–a₂ = 111–11 = 100

d₃ = a₄–a₃ = 1111–111 = 1000

⇒ As in A.P the difference between the 2 terms is always constant

But the difference in sequence is not the same.

∴ The above sequence is not A.P

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 5;

d₁ = a₂–a₁ = 15–5 = 10

d₂ = a₃–a₂ = 25–15 = 10

d₃ = a₄–a₃ = 35–25 = 10

⇒ As in A.P the difference between the 2 terms is always constant

The difference in sequence is same and comes to be 10.

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = 5 + (n–1)10

= 5 + 10n–10

= –5 + 10n

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 17;

d₁ = a₂–a₁ = 22–17 = 5

d₂ = a₃–a₂ = 27–22 = 5

d₃ = a₄–a₃ = 32–27 = 5

⇒ As in A.P the difference between the 2 terms is always constant

The difference in sequence is same and comes to be 5.

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = 17 + (n–1)5

= 17 + 5n–5

= 12 + 5n

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 101;

d₁ = a₂–a₁ = 99–101 = –2

d₂ = a₃–a₂ = 97–99 = –2

d₃ = a₄–a₃ = 95–97 = –2

⇒ As in A.P the difference between the 2 terms is always constant

The difference in sequence is same and comes to be (–2).

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = 101 + (n–1)(–2)

= 101–2n + 2

= 103–2n

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 201;

d₁ = a₂–a₁ = 198–201 = –3

d₂ = a₃–a₂ = 195–198 = –3

d₃ = a₄–a₃ = 192–195 = –3

⇒ As in A.P the difference between the 2 terms is always constant

The difference in sequence is same and comes to be (–3).

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = 201 + (n–1)(–3)

= 201–3n + 3

= 204–3n

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

The sequence of natural numbers which are consecutive multiples of 5

Is 5, 10, 15, 20 ……

In the above sequence,

a = 5;

d₁ = a₂–a₁ = 10–5 = 5

d₂ = a₃–a₂ = 15–10 = 5

d₃ = a₄–a₃ = 20–15 = 5

⇒ As in A.P the difference between the 2 terms is always constant

The difference in sequence is same and comes to be 5.

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = 5 + (n–1)5

= 5 + 5n–5

= 5n

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

The sequence of natural numbers which are multiple of 3 or 5

in increasing order

Is 3, 5, 6, 9, 10 ……

In the above sequence,

a = 3;

d₁ = a₂–a₁ = 5–3 = 2

d₂ = a₃–a₂ = 6–5 = 1

d₃ = a₄–a₃ = 9–6 = 3

⇒ As in A.P the difference between the 2 terms is always constant

The difference in sequence is not same .

∴ The above sequence is not A.P

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 2;

d₁ = a₂–a₁ = 7–2 = 5

d₂ = a₃–a₂ = 12–7 = 5

d₃ = a₄–a₃ = 17–12 = 5

The difference in sequence is same and comes to be 5 .

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = 2 + (n–1)(5)

= 2 + 5n–5

= –3 + 5n

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 200;

d₁ = a₂–a₁ = 195–200 = –5

d₂ = a₃–a₂ = 190–195 = –5

d₃ = a₄–a₃ = 185–190 = –5

The difference in sequence is same and comes to be (–5).

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = 200 + (n–1)(–5)

= 200–5n + 5

= 205–5n

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 1000;

d₁ = a₂–a₁ = 900–1000 = –100

d₂ = a₃–a₂ = 800–900 = –100

The difference in sequence is same and comes to be (–100).

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = 1000 + (n–1)(–100)

= 1000–100n + 100

= 1100–100n

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 50;

d₁ = a₂–a₁ = 100–50 = 50

d₂ = a₃–a₂ = 150–100 = 50

d₃ = a₄–a₃ = 200–150 = 50

The difference in sequence is same and comes to be 50.

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = 50 + (n–1)(50)

= 50 + 50n–50

= 50n

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = ;

d₁ = a₂–a₁ = = 1

d₂ = a₃–a₂ = = 1

d₃ = a₄–a₃ = = 1

The difference in sequence is same and comes to be (1).

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = + (n–1)(1)

= + n–1

= n–

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 1.1;

d₁ = a₂–a₁ = 2.1–1.1 = 1

d₂ = a₃–a₂ = 3.1–2.1 = 1

d₃ = a₄–a₃ = 4.1–3.1 = 1

The difference in sequence is same and comes to be (1).

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = 1.1 + (n–1)(1)

= 1.1 + n–1

= 0.1 + n

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 1.2;

d₁ = a₂–a₁ = 2.3–1.2 = 1.1

d₂ = a₃–a₂ = 3.4–2.3 = 1.1

d₃ = a₄–a₃ = 4.5–3.4 = 1.1

The difference in sequence is same and comes to be (1.1).

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = 1.2 + (n–1)(1.1)

= 1.2 + 1.1n–1.1

= 0.1 + 1.1n

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = ;

d₁ = a₂–a₁ = =

d₂ = a₃–a₂ = =

d₃ = a₄–a₃ = =

The difference in sequence is same and comes to be ().

∴ The above sequence is A.P

The n^th term of A.P is a_n = a + (n–1)d

a_n = a + (n–1)d = + (n–1)()

=

= 1 + n

Formula Used.

a_n = a + (n–1)d

a₇ = a + (7–1)d

a₇ = a + 6d

If 7^th term of A.P is given as 12

Then,

a + 6d = 12

we get a = 12–6d ......eq 1

a₁₂ = a + (12–1)d

a₁₂ = a + 11d

If 12^th term of A.P is given as 72

Then,

a + 11d = 72

we get a = 72–11d ......eq 2

Equating both eq 1 and eq 2

We get ;

12–6d = 72–11d

11d–6d = 72–12

5d = 60

d = = 12

Putting d in eq 1 we get ;

a = 12–6×12

= 12–72 = –60

As a = –60 and d = 12

Then;

a₁ = a + (n–1)d = –60 + (1–1)(12) = –60

a₂ = a + (n–1)d = –60 + (2–1)(12) = –60 + (12)×1 = –48

a₃ = a + (n–1)d = –60 + (3–1)(12) = –60 + (12)×2 = –36

a₄ = a + (n–1)d = –60 + (4–1)(12) = –60 + (12)×3 = –24

a_n = a + (n–1)d = –60 + (n–1)(12) = 12n – 72

∴ The A.P is –60, –48, –36, –24……, 12n – 72

Formula Used.

a_n = a + (n–1)d

a₂ = a + (2–1)d

a₂ = a + d

If 2^nd term of A.P is given as 1

Then,

a + d = 1

we get a = 1–d ......eq 1

a₁₂ = a + (12–1)d

a₁₂ = a + 11d

If 12^th term of A.P is given as –9

Then,

a + 11d = –9

we get a = –9–11d ......eq 2

Equating both eq 1 and eq 2

We get ;

1–d = –9–11d

11d–d = –9–1

10d = –10

d = = –1

Putting d in eq 1 we get ;

a = 1–(–1)

= 2

As a = 2 and d = –1

Then;

a₁ = a + (n–1)d = 2 + (1–1)(–1) = 2

a₂ = a + (n–1)d = 2 + (2–1)(–1) = 2 + (–1)×1 = 2–1 = 1

a₃ = a + (n–1)d = 2 + (3–1)(–1) = 2 + (–1)×2 = 2–2 = 0

a₄ = a + (n–1)d = 2 + (4–1)(–1) = 2 + (–1)×3 = 2–3 = –1

a_n = a + (n–1)d = 2 + (n–1)(–1) = 2 + 1–n = 3–n

∴ The A.P is 2, 1, 0, –1……, 3–n

Formula Used.

a_n = a + (n–1)d

a₃ = a + (3–1)d

a₃ = a + 2d

If 3^rd term of A.P is given as 8

Then,

a + 2d = 8

we get a = 8–2d ......eq 1

a₁₀ = a + (10–1)d

a₁₀ = a + 9d

If 10^th term of A.P is given as 20 + 6^th term

Then,

a₆ = a + (6–1)d

= a + 5d

a + 9d = 20 + [a + 5d]

we get

a–a + 9d–5d = 20

4d = 20

d = = 5 ......eq 2

Putting d in eq 1 we get ;

a = 8–(2×5)

= 8–10 = –2

As a = –2 and d = 5

Then;

a₁ = a + (n–1)d = –2 + (1–1)(5) = –2

a₂ = a + (n–1)d = –2 + (2–1)(5) = –2 + (5)×1 = –2 + 5 = 3

a₃ = a + (n–1)d = –2 + (3–1)(5) = –2 + (5)×2 = –2 + 10 = 8

a₄ = a + (n–1)d = –2 + (4–1)(5) = –2 + (5)×3 = –2 + 15 = 13

a_n = a + (n–1)d = –2 + (n–1)(5) = –2–5 + 5n = –7 + 5n

∴ The A.P is –2, 3, 8, 13, ……, 5n–7

Formula Used.

a_n = a + (n–1)d

a₅ = a + (5–1)d

a₅ = a + 4d

If 5^th term of A.P is given as 17

Then,

a + 4d = 17

we get a = 17–4d ......eq 1

a₉ = a + (9–1)d

a₉ = a + 8d

If 9^th term of A.P is given as 35 + 2^nd term

Then,

a₂ = a + (1)d

= a + d

a + 8d = 35 + [a + d]

we get

a–a + 8d–d = 35

7d = 35

d = = 5 ......eq 2

Putting d in eq 1 we get ;

a = 17–(4×5)

= 17–20 = –3

As a = –3 and d = 5

Then;

a₁ = a + (n–1)d = –3 + (1–1)(5) = –3

a₂ = a + (n–1)d = –3 + (2–1)(5) = –3 + (5)×1 = –3 + 5 = 2

a₃ = a + (n–1)d = –3 + (3–1)(5) = –3 + (5)×2 = –3 + 10 = 7

a₄ = a + (n–1)d = –3 + (4–1)(5) = –3 + (5)×3 = –3 + 15 = 12

a_n = a + (n–1)d = –3 + (n–1)(5) = –3 + 5n–5 = 5n–8

∴ The A.P is –3, 2, 7, 12, ……, 5n–8

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 12;

d₁ = a₂–a₁ = 17–12 = 5

d₂ = a₃–a₂ = 22–17 = 5

d₃ = a₄–a₃ = 27–22 = 5

The difference in sequence is same and comes to be (5).

For any term of A.P to be 0

a_n = a + (n–1)d = 0

a_n = a + (n–1)d = 12 + (n–1)(5)

= 12 + 5n–5

= 7 + 5n

7 + 5n = 0

5n = –7

n =

∴ The number of terms cannot be negative

Hence, the no term of A.P can be 0

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 201;

d₁ = a₂–a₁ = 197–201 = –4

d₂ = a₃–a₂ = 193–197 = –4

The difference in sequence is same and comes to be (–4).

For any term of A.P to be 5

a_n = a + (n–1)d = 5

a_n = a + (n–1)d = 201 + (n–1)(–4)

5 = 201–4n + 4

5 = 205–4n

–4n = –205 + 5

n = = 50

∴ The 50^th term of A.P is 5

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 8;

d₁ = a₂–a₁ = 11–8 = 3

d₂ = a₃–a₂ = 14–11 = 3

d₃ = a₄–a₃ = 17–14 = 3

The difference in sequence is same and comes to be (3).

For any term of A.P to be 272

a_n = a + (n–1)d = 272

a_n = a + (n–1)d = 8 + (n–1)(3)

= 8 + 3n–3

= 5 + 3n

5 + 3n = 272

3n = 272–5 = 267

n = = 89

∴ The 89^th term of A.P has value 272

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 3;

d₁ = a₂–a₁ = 6–3 = 3

d₂ = a₃–a₂ = 9–6 = 3

d₃ = a₄–a₃ = 12–9 = 3

The difference in sequence is same and comes to be (3).

As the last term is 300

a_n = a + (n–1)d

300 = 3 + (n–1)3

300 = 3 + 3n–3 = 3n

n = = 100^th term

10^th term from the end is

100 + 1–10 = 91^st term

∵ while counting, last term is also counted

a_n = a + (n–1)d

a₉₁ = 3 + (91–1)3

= 3 + 90×3

= 273

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 10;

d₁ = a₂–a₁ = 15–10 = 5

d₂ = a₃–a₂ = 20–15 = 5

d₃ = a₄–a₃ = 25–20 = 5

The difference in sequence is same and comes to be (5).

As the last term is 1000

a_n = a + (n–1)d

1000 = 10 + (n–1)5

1000 = 10 + 5n–5 = 5 + 5n

5n = 995

n = = 199^th term

15 term from the end will be

199 + 1–15 = 185^th term

a_n = a + (n–1)d

a₁₈₅ = 10 + (185–1)5

= 10 + 184×5

= 10 + 920

= 930

Formula Used.

a_n = a + (n–1)d

a₇ = a + (7–1)d

a₇ = a + 6d

If 7^th term of A.P is given as 18

Then,

a + 6d = 18

we get a = 18–6d ......eq 1

a₁₈ = a + (18–1)d

a₁₈ = a + 17d

If 18^th term of A.P is given as 7

Then,

a + 17d = 7

we get a = 7–17d ......eq 2

Equating both eq 1 and eq 2

We get ;

18–6d = 7–17d

17d–6d = 7–18

11d = –11

d = = –1

Putting d in eq 1 we get ;

a = 18–6×(–1)

= 18 + 6 = 24

For 101^st term

a_n = a + (n–1)d

a₁₀₁ = 24 + (101–1)(–1)

= 24–100

= –76

Formula Used.

a_n = a + (n–1)d

a_m = a + (m–1)d

If m^th term of A.P is given as n

Then,

a + (m–1)d = n

we get a = n–(m–1)d ......eq 1

a_n = a + (n–1)d

If n^th term of A.P is given as m

Then,

a + (n–1)d = m

we get a = m–(n–1)d ......eq 2

Equating both eq 1 and eq 2

We get ;

n–(m–1)d = m–(n–1)d

(n–1)d–(m–1)d = m–n

(n–1–(m–1))d = m–n

(n–1–m + 1)d = m–n

(n–m)d = –(n–m)

d = = –1

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the following A.P

a = 2,

d = a₂–a₁

= 6–2 = 4;

n = 20

S_n = [2a + (n–1)d]

= [2×2 + (20–1)4]

= 10[4 + 19×4]

= 10[4 + 76]

= 800

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the following A.P

a = 5,

d = a₂–a₁

= 7–5 = 2;

n = 30

S_n = [2a + (n–1)d]

= [2×5 + (30–1)2]

= 15[10 + 29×2]

= 15[10 + 58]

= 1020

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the following A.P

a = –10,

d = a₂–a₁

= –12–(–10) = (–2);

n = 15

S_n = [2a + (n–1)d]

= [2×(–10) + (15–1)(–2)]

= [–20 + 14×(–2)]

= [–20–28]

= 15×(–24)

= –360

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the following A.P

a = 1,

d = a₂–a₁

= 1.5–1 = 0.5;

n = 16

S_n = [2a + (n–1)d]

= [2×1 + (16–1)0.5]

= 8[2 + 15×0.5]

= 8[2 + 7.5]

= 76

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the following A.P

a = ,

d = a₂–a₁

= = 1;

n = 18

S_n = [2a + (n–1)d]

= [2× + (18–1)1]

= 9[ + 17]

= 9[]

= 159

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the sum above A.P follows

Is 3, 6, 9, ……, 300

In the following A.P

⇒ a = 3,

⇒ d = 6–3 = 3

As the last term is 300

a_n = a + (n–1)d

300 = 3 + (n–1)3

300 = 3 + 3n–3

3n = 300

⇒ n = = 100

S_n = [2a + (n–1)d]

= [2×3 + (100–1)3]

= 50[6 + 99×3]

= 50[6 + 297]

= 15150

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the sum above A.P follows

Is 5, 10, 15, ……, 100

In the following A.P

⇒ a = 5,

⇒ d = 10–5 = 5

As the last term is 100

a_n = a + (n–1)d

100 = 5 + (n–1)5

100 = 5 + 5n–5

5n = 100

⇒ n = = 20

S_n = [2a + (n–1)d]

= [2×5 + (20–1)5]

= 10[10 + 19×5]

= 10[10 + 95]

= 1050

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the sum above A.P follows

Is 7, 12, 17, 22, ……, 102

In the following A.P

⇒ a = 7,

⇒ d = 12–7 = 5

As the last term is 102

a_n = a + (n–1)d

102 = 7 + (n–1)5

102 = 7 + 5n–5

5n = 102–2

⇒ n = = 20

S_n = [2a + (n–1)d]

= [2×7 + (20–1)5]

= 10[14 + 19×5]

= 10[14 + 95]

= 1090

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the sum above A.P follows

Is –100, –92, –84, ……, 92

In the following A.P

⇒ a = –100,

⇒ d = –92–(–100) = –92 + 100 = 8

As the last term is 92

a_n = a + (n–1)d

92 = –100 + (n–1)8

92 = –100 + 8n–8

8n = 92 + 108

⇒ n = = 25

S_n = [2a + (n–1)d]

= [2×(–100) + (25–1)8]

= [–200 + 24×8]

= [–200 + 192]

= –100

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the sum above A.P follows

Is 25, 21, 17, 13, ……, (–51)

In the following A.P

⇒ a = 25,

⇒ d = 21–25 = –4

As the last term is –51

a_n = a + (n–1)d

–51 = 25 + (n–1)(–4)

–51 = 25–4n + 4

4n = 80

⇒ n = = 20

S_n = [2a + (n–1)d]

= [2×25 + (20–1)(–4)]

= 10[50 + 19×(–4)]

= 10[50–76]

= –260

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the following A.P

⇒ a = 1,

⇒ d = 2

⇒ n = 10

S_n = [2a + (n–1)d]

= [2×1 + (10–1)2]

= 5[2 + 9×2]

= 5[2 + 18]

= 100

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the following A.P

⇒ a = 2,

⇒ d = 3

⇒ n = 30

S_n = [2a + (n–1)d]

= [2×2 + (30–1)3]

= 15[4 + 29×3]

= 15[4 + 87]

= 1365

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

In the following A.P

⇒ S₃ = 9,

S₃ = [2a + (3–1)d]

9 = [2a + 2d]

9 = [2(a + d)]

= [a + d]

3 = [a + d]

⇒ a = 3–d ……eq 1

⇒ S₇ = 49

S₇ = [2a + (7–1)d]

49 = [2a + 6d]

49 = [2(a + 3d)]

= a + 3d

7 = a + 3d

⇒ a = 7–3d ......eq 2

By equating eq 1 and eq 2

3–d = 7–3d

3d – d = 7 – 3

2d = 4

d = = 2

Putting value of d in eq 1

a = 3–d = 3–2

= 1

S_n = [2a + (n–1)d]

= [2×1 + (n–1)2]

= [2 + 2n–2]

= ×2n

= n²

∴ S_n = n²

⇒ S₁₀ = (10)²

= 100

Formula used.

S_n = [a + a_n]

T_n = a_n = a + (n–1)d

S_n = [2a + (n–1)d]

⇒ S₁₀ = 320

⇒ T₁₀ = a₁₀ = 41

S₁₀ = [a + 41]

320 = [a + 41]

= a + 41

64 = a + 41

a = 64–41

⇒ a = 23

a₁₀ = a + (n–1)d

41 = 23 + (10–1)d

41–23 = 9d

9d = 18

⇒ d = = 2

T_n = a_n = a + (n–1)d

T_n = 23 + (n–1)2

23–2 + 2n

21 + 2n

S_n = [2a + (n–1)d]

S_n = [2×23 + (n–1)2]

= [46 + 2n–2]

= [44 + 2n]

= n[22 + n]

= 22n + n²

Formula used.

S_n = [2a + (n–1)d]

⇒ S₁₀ = 50

⇒ n = 10

⇒ a = 0.5S₁₀ = [2×0.5 + (10–1)d]

50 = 5[1 + 9d]

= 1 + 9d

9d = 10–1

d = = 1

Formula used.

S_n = [2a + (n–1)d]

⇒ S₂₀ = 100

⇒ n = 20

⇒ d = –2

S₂₀ = [2a + (20–1)(–2)]

100 = 10[2a–38]

100 = 20[a–19]

= a–19

5 = a–19

a = 19 + 5

a = 24

Formula used.

S_n = [2a + (n–1)d]

d = a_{n + 1}–a_n

⇒ S₂₀ = 990

⇒ a = 2

⇒ d = a_{n + 1}–a_n

= 7 – 2 = 5

S_n = [2×2 + (n–1)5]

990 = [4 + 5n–5]

990 = [5n–1]

990×2 = 5n²–n

5n²–n–1980 = 0

n =

∴ n can be 20 or –19.8

Since, the number of terms cannot be negative

∴ Sum of 20 terms gives the sum of 990

Formula used.

S_n = [a + l]

⇒ S_n = 500

⇒ a = 5

⇒ l = 45

S_n = [a + l]

500 = [5 + 45]

500×2 = n×50

n = = 20

∴ Number of terms to get sum 500 is 20

Formula used.

a_n = a + (n–1)d

S_n = [2a + (n–1)d]

⇒ d = 5

⇒ a = 5

⇒ l = 95

a_n = a + (n–1)d

95 = 5 + (n–1)5

95 = 5–5 + 5n

n = = 19

S_n = [a + l]

= [5 + 95]

= 19×50

= 950

∴ Number of terms is 19 and Sum of A.P is 950

Formula used.

S_n = [2a + (n–1)d]

S_n = [2a + (n–1)d]

5n – 2n² = [2a + (n–1)d]

2×n[5 – 2n] = n[2a + (n–1)d]

10 – 4n = (2a–d) + nd

Then

Real part should be equal to Real

And imaginary part should be equal to imaginary

2a – d = 10 ......eq 1

nd = –4n

⇒ d = –4

Putting value of d in eq 1

2a – d = 10

2a – (–4) = 10

2a + 4 = 10

2a = 10 – 4 = 6

⇒ a = = 3

Formula used.

S_n = [a + l]

a_n = a + (n–1)d

⇒ The 1^st 3 digit number divisible by 3 is 102

For the last 3 digit number divisible by 3

Divide 999 by 3

It gets completely divisible by 3

⇒ The last 3 digit number divisible by 3 is 999

a_n = a + (n–1)d

999 = 102 + (n–1)3

999 = 102–3 + 3n

3n = 999 – 99

n = = 300

S_n = [a + l]

= [102 + 999]

= 150×1101

= 165150

Formula used.

S_n = [a + l]

a_n = a + (n–1)d

⇒ The 1^st odd number is 5

⇒ The last odd number is 205

⇒ Difference between every odd number is 2

a_n = a + (n–1)d

205 = 5 + (n–1)2

205 = 5–2 + 2n

2n = 205 – 3

n = = 101

S_n = [a + l]

= [5 + 205]

= 101×105

= 10605

Formula used.

S_n = [2a + (n–1)d]

a_n = a + (n–1)d

d = a_{n + 1} – a_n

Let the 1^st negative term be X

a = 121

d = 117 – 121

= –4

If the difference is –4 then

Then X can be (–1), (–2), (–3), (–4)

a_n = a + (n–1)d

X = 121 + (n–1)(–4)

X = 125–4n

4n = 125 – X

n =

For the number of terms to be an integer (125–X) will be multiple of 4

When we put X = –1

n will be

which is not an integer

When we put X = –2

n will be

which is not an integer

When we put X = –3

n will be = 32

which is an integer

Hence –3 is the 1^st negative term.

S_n = [2a + (n–1)d]

= [2×121 + (32–1)(–4)]

= 16[242 – 124]

= 16×118

= 1888

From the question we know that, T_n = 6n + 5 ……. (1)

And we know that T_n = a + (n – 1)d

So we have,

⇒ a + (n – 1)d = 6n + 5

⇒ a + (n – 1)d = 6n + 11 – 6

⇒ a + (n – 1)d = 11 + (6n – 6)

⇒ a + (n – 1)d = 11 + 6(n – 1)

On comparing both the sides, we have:

⇒ a = 11 and d = 6

So now, S_n = × (2a + (n – 1)d)

⇒ S_n = × (2(11) + (n – 1)(6))

⇒ S_n = × (22 + 6n – 6)

⇒ S_n = × (16 + 6n)

⇒ S_n = n × (8 + 3n)

⇒ S_n = 3n² + 8n

∴ Sum of n terms of the given A.P is S_n = 3n² + 8n

From the question we know that, S_n = n² + 2n ……. (1)

We know that, S_n = × (2a + (n – 1)d)

= na +

= na + (n² – n) ×

S_n = na + n² × – n × ……. (2)

As we know that, (1) and (2) are both sum of the same arithmetic progression

So we can equate them to each other.

So, we have,

⇒ n² + 2n = na + n² × – n ×

⇒ n² + 2n = n² × + na – n ×

Now we will equate the coefficients of “n²” and “n” on both the sides of the equal to sign.

So, we get,

For coefficients of n² :

⇒ 1 =

⇒ d=2

Now, For coefficients of n:

⇒ 2 = a –

From above we have that d=2,

So we can say that,

⇒ 2 = a –

⇒ 2 = a – 1

⇒ a=3

So, now, we know that, T_n = a + (n – 1)d

Now we put values of a and n in the above equation,

We get,

⇒ T_n = 3 + (n – 1)2

⇒ T_n = 3 + 2n – 2

⇒ T_n = 2n + 1

∴ for the given value of S_n = n² + 2n, we have T_n = 2n + 1.

We can see that, A.P. is 30, 27, 24, 21 …

T₁ = 30 = a

T₂ = 27

So, we have d = T₂ – T₁

d = 27 – 30

d= – 3

Now, we have sum of the A.P. S_n = × (2a + (n – 1)d)

And it is given that, S_n = 120

So, we have,

⇒ 120 = × (2a + (n – 1)d)

Now we will put the values of a and d in the above equation

⇒ 120 = × (2(30) + (n – 1)(– 3))

⇒ 120 = × (60 + 3 – 3n)

120 × 2 = n × (63 – 3n)

⇒ 240 = 63n – 3n²

⇒ 3n² – 63n + 240 = 0

⇒ n² – 21n + 80 = 0

⇒ n² – 16n – 5n + 80 = 0

⇒ n(n – 16) – 5(n – 16) = 0

⇒ (n – 16)(n – 5) = 0

⇒ n= 16 or 5

Now, when n=16, the last term = T₁₆

So we have, T₁₆ = a + (16 – 1)d ……. (∵ T_n = a + (n – 1)d )

⇒ T₁₆ = a + 15d

Now we put the values of a and d in the above equation

⇒ T₁₆= 30 + 15(– 3)

⇒ T₁₆ = 30 – 45

⇒ T₁₆ = – 15

Now, when n=5, the last term = T₅

So we have, T5 = a + (5 – 1)d ……. (∵ Tn = a + (n – 1)d )

⇒ T₅ = a + 4d

Now we put the values of a and d in the above equation

⇒ T₅= 30 + 4(– 3)

⇒ T₅ = 30 – 12

⇒ T₅ = 18

∴ The no. Of term in the above A.P. can be n = 5 or 16

And the last terms of the respective A.Ps are:

For n = 5, last term T5 = 18

For n = 16, last term T16 = – 15

We have the A.P as 100, 97, 94, 91 …

So, a = 100

⇒ d= 97 – 100 = – 3

Now recall that, nth term of an AP is T_n = a + (n – 1)d

⇒ T_n = 100 + (n – 1)(– 3)

= 100 – 3n + 3

⇒ T_n = 103 – 3n

Since we need negative term, T_n < 0

So, 103 – 3n < 0

⇒ 103 < 3n

⇒ < n

⇒ 34.333 < n

⇒ n>34

i.e. N = 35

∴ n=35 for the first negative term in the given A.P.

⇒ T₃₅ = a + (35 – 1)d

= 100 + 34 × – 3

= 100 – 102

= – 2

∴ the first negative term of the given AP is – 2.

Now the first 3 digit multiple of 6 is 102.

That means, a = 102

And all the nos. Are multiples of 6 which means that they have a difference of 6 in between them so, common difference d = 6

Now, the last 3 digit multiple of 6 is 996.

Now, we know that, T_n = a + (n – 1)d

So, 996 = a + (n – 1)d

⇒ 996 = 102 + (n – 1)6

⇒ 6(n – 1) = 996 – 102

⇒ 6(n – 1) = 894

⇒ n – 1 =

⇒ n – 1 = 149

⇒ n = 150

So, we can say that there are 150, 3 digit multiples of 6

Now sum of these nos. S₁₅₀ = × (2a + (n – 1)d)

⇒ S₁₅₀ = × (2(102) + (150 – 1)6)

= 75 × (204 + 149 × 6)

= 75 × (1098)

⇒ S₁₅₀ = 82350

∴ Sum of all 3 digit natural multiples of 6 is 82350.

As we know that sum of n terms of an A.P is:

⇒ S_n= × (2a + (n – 1)d)

So, sum of m terms of an A.P is:

⇒ S_m = × (2a + (m – 1)d)

From the question we know that,

⇒ =

⇒ =

⇒ =

⇒ =

⇒ m × (2a + (n – 1)d) = n × (2a + (m – 1)d)

⇒ 2ma + m(n – 1)d = 2na + n(m – 1)d

⇒ 2ma + md(n – 1) = 2na + nd(m – 1)

⇒ 2ma + mnd – md = 2na + mnd – nd

⇒ 2ma – md = 2na – nd

⇒ 2ma – 2na = md – nd

⇒ 2a(m – n) = d(m – n)

⇒ 2a = d

Now, m^th term of the given A.P is T_m = a + (m – 1)d

Now, n^th term of the given A.P is T_n = a + (n – 1)d

Now ratio between these 2 terms is:

⇒ =

Now we have, 2a = d

So we get,

⇒ =

⇒ =

⇒ =

⇒ =

⇒ =

∴ the ratio of m^th term of the given A.P to its n^th Term is:

T_m : T_n = 2m – 1 : 2n – 1

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

Sum of the first p terms is:

S_l = × (2a + (l – 1)d) = p ………(1)

Sum of the first m terms is:

⇒ S_m = × (2a + (m – 1)d) = q ……… (2)

Sum of the first n terms is:

⇒ S_n = × (2a + (n – 1)d) = r ……… (3)

Now, (1) × + (2) × + (3) ×

We get, p × + q × + r ×

= 0 + [(m – n)l + (n – l)m + (l – m)n –m + n – n + l – l + m]

= [lm – ln + mn – lm + ln – mn + 0]

= × 0 = 0

∴ it is proved that, p × + q × + r × = 0

Let the sum of n terms of the first A.P be:

⇒ S_n = × (2a + (n – 1)d) …………… (1)

Let the sum of n terms of the second A.P be:

⇒ S’_n = × (2a’ + (n – 1)d’) …………… (2)

Now according to the question:

⇒ =

Let’s consider the ratio these two AP’s m^th terms as:

T_m : T’_m

Now, recall that, nth term of an AP is T_n = a + (n – 1)d

⇒ T_m = a + (m – 1)d

⇒ T’^m = a’ + (m – 1)d’

Hence the ratio of these two AP’s m^th terms become:

⇒ =

On multiplying by 2, we get,

⇒ =

⇒ =

⇒ =

⇒ =

⇒ =

⇒ =

Now from the above formula of the ratio of mth terms of 2 Aps, we can find the ratio of 7^th terms of both Aps

So we have, =

⇒ =

⇒ =

∴ the ratio of m^th terms of the given 2 Aps is, 16m – 7 : 14m – 4

∴ the ratio of 7^th terms of the given 2 Aps is, 105 : 94

We know that the sum of 3 numbers in the AP is 18.

Let’s say that those 3 numbers are: a – d, a, a + d

So we can say that, a – d + a + a + d = 18

⇒ 3a = 18

⇒ a= 6

Now, we have that the sum of squares of these 3 numbers is 180.

So we can say that, (a – d) ² + a² + (a + d) ² = 180

⇒ a² – 2ad + d² + a² + a² + 2ad + d² = 180

⇒ 3a² + 2d² = 180

We know that a = 6,

So, 3(6)² + 2d² = 180

⇒ 3(36) + 2d² = 180

⇒ 108 + 2d² = 180

⇒ 2d² = 180 – 108

⇒ 2d² = 72

⇒ d² = 36

⇒ d = 6

Now, our 3 numbers a – d, a, a + d are 6 – 6, 6, 6 + 6 = 0, 6, 12 respectively.

∴ The 3 numbers of the A.P in the increasing order are : 0, 6, 12.

From the given data we calculate the distance covered for each potato.

So, for first potato, distance = 2 × 5 = 10m

For second, distance = 10 + 2 × 3= 16m

For third, distance = 16 + 2 × 3= 22m

Thus the distance to be covered form an AP. :

10, 16, 22, 28, …………

The total distance to be covered for 15 potatoes is given by S₁₅ .

Hence, for the above AP, we have a = 10 and d = 6.

So, we know that, S_n = × (2a + (n – 1)d)

⇒ S₁₅= × [2(10) + (15 – 1) × 6]

= × [20 + 14 × 6]

= × [20 + 84]

= × 104

= 15 × 52

= 780m

∴ if 15 potatoes are placed in the race, the total distance covered is 780m.

Now, it is given that, the total distance covered is 1340m and we need to find the no. Of potatoes.

So, let the no. Of potatoes be n, then we take,

⇒ S_n = 1340

⇒ 1340 = × (2a + (n – 1)d)

⇒ 1340 = × (2(10) + (n – 1)(6))

⇒ 1340 = × (20 + 6n – 6)

⇒ 1340 = × (14 + 6n)

⇒ 1340 = n × (7 + 3n)

⇒ 3n² + 7n – 1340 = 0

On solving we get that,

⇒ n = or n = 20

as we n cannot be negative, so we have n = 20.

∴ if total distance to be covered is 1340m, then no. Of potatoes placed in the race are 20.

The distance between 2 consecutive rungs is 25cm and the distance between the top and bottom rung is 2.5m = 250cm.

∴ no. Of rungs = + 1 = 11

The length of the bottom rung is 60cm and going upwards the length of the rung decreases uniformly.

The length of last rung is 40cm.

So, the length of the rung will form a finite AP.

With first term a = 60(T₁) and the last 11^th term as 40(T₁₁)

Now, d =

=

=

⇒ d= – 2

the length of the wood required is given by S₁₁

we know that, S_n = × (2a + (n – 1)d)

So, S₁₁ = × (2a + (n – 1)d)

⇒ S₁₁ = × (2(60) + (11 – 1)( – 2))

⇒ S₁₁ = × ( 120 – 20)

⇒ S₁₁ = × 100

⇒ S₁₁ = 550

∴ the length of the wood required is 550cm i.e. 5.5m.

The mount paid as down payment is 200Rs.

Amount paid in 1^st instalment = 200 + 150 = 350Rs

Amount paid in 2^nd instalment = 350 + 150 = 500Rs and so on.

The amount paid every month increase every month and forms finite AP. Which is : 200, 350, 500 ……..

Total amount paid is S_n = 32, 500Rs

we know that, S_n = × (2a + (n – 1)d)

⇒ 32, 500 = × (2(200) + (n – 1)(150))

⇒ 32, 500 = × (400 + 150n – 150)

⇒ 32, 500 = × (250 + 150n)

⇒ 32, 500 =n × (125 + 75n)

⇒ 75n² + 125n – 32500 = 0

⇒ 25n² + 5n – 1300 = 0

⇒ (3n + 65)(n – 20)=0

⇒ n = or n = 20

As n cannot be negative, so we have n = 20.

Thus we can say that there are 20 terms in the AP.

Among these terms first payment is the down payment, so man took 20 – 1 = 19 months to complete the payment.

∴ man took 19 months to complete the payment.

We know that, T_n = a + (n – 1)d

So for T_1,

We have T₁=a + (1 – 1)d = a

According to the question, a = 22

Now, we have T_n = – 11

Ie., a + (n – 1)d = – 11

So, 22 + (n – 1)d = – 11

⇒ (n – 1)d = – 22 – 11

⇒ (n – 1)d = – 33

Now we have that, S_n = × (2a + (n – 1)d)

From the question we can say that, S_n = 66

So, we have,

⇒ 66= × (2a + (n – 1)d)

⇒ 132 = n × (2a + (n – 1)d)

We have (n – 1)d = – 33 and a = 22, so we put that in the above equation.

⇒ 132 = n × (2(22) – 33)

⇒ 132 = n × (44 – 33)

⇒ 132= n × 11

⇒ n =

⇒ n = 12

∴ value of n is 12.

We have a = 8

We know that, T_n = a + (n – 1)d

And from the question we can say that, T_n= 53

So we get,

⇒ 33 = a + (n – 1)d

Putting value of a in the above equation,

⇒ 33 = 8 + (n – 1)d

⇒ (n – 1)d = 33 – 8

⇒ (n – 1)d = 25

Now we have that, S_n = × (2a + (n – 1)d)

From the question we can say that, S_n = 123

So, we have,

⇒ 123 = × (2a + (n – 1)d)

⇒ 246 = n × (2a + (n – 1)d)

now we have a = 8 and (n – 1)d = 25, so we put them in the above equation,

⇒ 246 = n × (2(8) + 25)

⇒ 246 = n × (16 + 25)

⇒ 246 = n × (31)

⇒ n= = 6

⇒ n = 6

now, we know that, (n – 1)d = 25

we put the value of n in the above equation, we get

⇒ (6 – 1)d = 25

⇒ 5d = 25

⇒ d = 5

∴ for the given A.P, value of n is 6 and value of d is 5.

We have, T₃ = 8

And recall that, T_n = a + (n – 1)d

So we have, T₃ = a + (3 – 1)d

⇒ 8 = a + 2d ……(1)

We also have, T₇ = 24

And T₇ = a + (7 – 1)d

⇒ 24 = a + 6d …… (2)

From (1), we have that, a = 8 – 2d

So we put the value of a in (2)

⇒ 24 = 8 – 2d + 6d

⇒ 24 – 8 = 4d

⇒ 16 = 4d

⇒ d= 4

now, putting value of d in (1),

⇒ 8 = a + 2(4)

⇒ 8 = a + 8

⇒ a =0

now, T₁₀= a + (10 – 1)d

⇒ T₁₀= 0 + 9 × 4

⇒ T₁₀= 36

∴ correct option is (d).

We have S_n = 2n² + 3n from the question.

We know that, S_n = × (2a + (n – 1)d)

So we can say that,

⇒ 2n² + 3n = × (2a + (n – 1)d)

⇒ 4n² + 6n = n × (2a + (n – 1)d)

⇒ 4n² + 6n = 2na + dn² – dn

⇒ 4n² + 6n = dn² + (2a – d)n

Now comparing the coefficients of “n^2” and “n”, we get that,

d = 4

∴ the correct option is (b).

We know that the sum of 3 nos in the AP is 48.

Lets say that those 3 nos are : a – d, a, a + d

So we can say that, a – d + a + a + d = 48

3a = 48

a= 16

now we have product of the first and the last term out of these 3 to be 252.

Ie. (a – d)(a + d) = 252

⇒ a² – d² = 252

we have a = 16, so we put the value of a in the above equation:

⇒ (16)² – d² = 252

⇒ 256 – d² = 252

⇒ d² = 256 – 252

⇒ d² = 4

⇒ d=2

∴ the correct option is (a)

We have a = 2

d= 4

n = 20

we also know that, S_n = × (2a + (n – 1)d)

⇒ S₂₀ = × (2(2) + (20 – 1)4)

⇒ S₂₀ = 10 × (4 + 19 × 4)

⇒ S₂₀ = 10 × (4 + 76)

⇒ S₂₀ = 10 × 80

⇒ S₂₀ = 800

∴ the correct option is (b)

We have a = 3

Then d = 5 – 3 = 2

Then, S_n = 288

We can recall that, S_n = × (2a + (n – 1)d)

So, S_n = × (2(3) + (n – 1)2)

⇒ S_n = × (6 + 2n – 2)

⇒ S_n = × (4 + 2n)

⇒ S_n = n × (2 + n)

⇒ S_n = n² + 2n

⇒ n² + 2n = 288

⇒ n² + 2n – 288 = 0

⇒ n² + 18n – 16n – 288 = 0

⇒ n(n + 18) – 16(n + 18) = 0

⇒ (n + 18)(n – 16)= 0

So we have n = – 18 or 16

But as n cannot be negative so, we have n = 16

∴ the correct option is (c)

We have n = 4

⇒ S₄ = 72

Let the 4 nos. Be, a, a + d, a + 2d, a + 3d

Now we have that a + a + d + a + 2d + a + 3d = 72

⇒ 4a + 6d = 72

⇒ 2a + 3d = 36 …….(1)

Now We know that the largest of them is twice the smallest.

So we can say that,

⇒ 2a = a + 3d

⇒ a = 3d

we now put the value of a in (1),

to get :

⇒ 2a + a = 36

⇒ 3a = 36

⇒ a = 12

we know that a is the first term in this AP, so the possible option is (b)

but to check whether it is the correct we will have to check the relation between the first and the last term of the AP of (b) option.

We can see that, in (b), first term is 12

And the last term is 24 so, this is the correct AP.

∴ the correct option is (b)

We know that, S_n = × (a + l)

Where a = first term of the AP.

And l = last term of the AP

For S_1,

We have S₁ = × (2 + 2n)

⇒ S₁ = n × (n + 1)

For S_2,

We have S₂ = × (1 + 2n – 1)

⇒ S₂ = × (2n)

⇒ S₂ = n²

Now = =

∴ =

∴ the correct option is (a)

We can recall that, S_n = × (2a + (n – 1)d) ……(1)

So now, S_{n – 1} = × (2a + ((n – 1) – 1)d)

⇒ S_{n – 1} = × (2a + (n – 2)d) …….(2)

So now, S_{n – 2} = × (2a + ((n – 2) – 1)d)

⇒ S_{n – 2} = × (2a + (n – 3)d) ………(3)

Now putting the above values in the equation_,

We get,

⇒ S_n — 2S_{n – 1} + S_{n – 2}

= × (2a + (n – 1)d) – 2[ × (2a + (n – 2)d)] + × (2a + (n – 3)d)

= – 2[] +

= an + – – 2an – dn² + 2a – 2d + 3nd + an + – – 2a + 3d

= [an – 2an + an] + [ – dn² + ] + [2a – 2a] + [ – + 3nd – ] + [ – 2d + 3d]

= 0 + 0 + 0 + 0 + d

= d

∴ S_n — 2S_{n – 1} + S_{n – 2} = d

∴ the correct option is (b)

⇒ S_m = n = × [ 2a + (m – 1)d ]

⇒ = 2a + (m – 1)d ......................(1)

⇒ S_n = m = × [ 2a + (n – 1)d ]

⇒ = 2a + (n – 1)d ........................(2)

subtracting both equations, we get :

⇒ 2( – ) = d(m – n)

⇒ d = – 2[] ...................(3)

now, S_{m + n} = × [ 2a + (m + n – 1)d ]

⇒ S_{m + n} × = 2a + (m + n – 1)d .......................(4)

now, (4) – (2), we get :

⇒ S_{m + n} × – = d(m)

putting value of d from (3), we get :

⇒ S_{m + n} × – = – 2[] × m

⇒ S_{m + n} × = – 2[] × m +

⇒ S_{m + n} × = – 2[] +

⇒ S_{m + n} × = – 2 or

⇒ S_{m + n} = – (m + n)

∴ the correct option is (a).

Given T₄ = 7

And we know that, T_n = a + (n – 1)d

So we have T₄ = a + (4 – 1)d

⇒ 7 = a + 3d …… (1)

Then, T₇ = 4

⇒ T₇ = a + (7 – 1)d

⇒ 4 = a + 6d …….(2)

Now, (2) – (1) gives :

⇒ – 3 = 3d

⇒ d = – 1

so, now, putting value of d in (1),

⇒ 7 = a + 3d

⇒ 7 = a + 3( – 1)

⇒ 7 = a – 3

⇒ a = 10

so now T₁₀ = a + (10 – 1)d

⇒ T₁₀ = a + 9d

= 10 + 9( – 1)

= 10 – 9

= 1

∴ T₁₀ = 1

∴ correct option is (d)

If 2k + 1, 13, 5k — 3 are three consecutive terms of A.P., then,

⇒ 13 – (2k + 1) = (5k – 3 ) – 13 …..(common difference)

⇒ 12 – 2k = 5k – 16

⇒ 28 = 7k

⇒ k = 4

∴ correct option is (c).

(1) + (1 + 1) + (1 + 1 + 1) + .…. + (1 + 1 + 1 + ...n — 1 times)

= 1 + 2 + 3 + 4 ………… + (n – 1)

= S_{n – 1}

⇒ S_{n – 1} = × [2a + (n – 1 – 1)d]

= × [2(1) + (n – 2)(1)]

=

=

=

=

∴ the correct option is (a).

If we see that, in the given series, d = 2

So if e continue the series, we get :

5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25 …….

We can see that, the prime nos. In the above are :

5, 7, 11, 13, 17, 19, 23, …

so here we see that the sixth term which is prime is 19

∴ the correct option is (b)

We know that, T_n = a + (n – 1)d

So we have, T₁₈= a + (18 – 1)d

⇒ T₁₈= a + 17d

Now T₈= a + (8 – 1)d

⇒ T₈= a + 7d

Now we have,

⇒ T₁₈ — T₈ = [a + 17d] – [a + 7d]

= 10d

∴ the correct option is (b)

We know that, T_n = a + (n – 1)d

So we have, T₂₅= a + (25 – 1)d

⇒ T₂₅= a + 24d

Now we have, T₂₀= a + (20 – 1)d

⇒ T₂₀= a + 19d

It is given that, T₂₅ — T₂₀ = 15

So we can say that,

⇒ 15 = [a + 24d] – [a + 19d]

⇒ 15 = 5d

⇒ d = 3

∴ the correct option is (a).