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Gievn that a_{1 }= 2k - 7, a_{2} = k + 5 and a_{3} = 3k + 2

d = a_{2} - a_{1} = k + 5 - (2k - 7) ............ (i)

d = a_{3} - a_{2} = 3k + 2 - (k + 5) ............. (ii)

From equation (i) and (ii)

k + 5 - (2k - 7) = 3k + 2 - (k + 5)

k + 5 - 2k + 7 = 3k + 2 - k - 5

- k + 12 = 2k - 3

2k + k = 12 + 3

3k = 15

k = 15/3

k = 5 **Answer**

Let the first term = a and common difference = d of the First A.P.

and the first term = a1 and common difference = d1 of the Second A.P.

Here we have to find ratio of 9^{th} term So n will be 2k -1 where k will be 9

Then n = 2k-1 = 2 x 9 - 1 = 17

Now putting the value n = 17 in above equation

Hence Ratio of 9th term = 59:134 Answer

Let the first term of A.P. is a and common difference d = d,

अत: m(a_{m}) = n(a_{n})

**⇒** m[a + (m – 1)d] = n[a + (n – 1)d]

**⇒** am + m(m – 1)d = an + n(n – 1)d

**⇒** am – an + [m^{2} – m – n^{2 }+ n]d = 0

**⇒** a(m – n ) + [m^{2} – n^{2 }–( m – n)]d = 0

**⇒** a(m – n ) + [(m + n) (m – n) – ( m – n)]d = 0

**⇒** (m – n ) [a + {(m + n)– 1}]d = 0

**⇒** a + {(m + n)– 1}d = 0 ………….. (1)

a_{m + n }= a + {(m + n)– 1}d

a_{m + n }= 0 From Equation (i)

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

a_{p} = a = A + (p -1)d

a = A + (p -1)d …………. (1)

a_{q} = b = A + (q -1)d

b = A + (q -1)d …………. (2)

a_{r} = c = A + (r -1)d

c = A + (r -1)d …………. (3)

a(q – r) + b(r – p) + c(p –q)

= {A + (p -1)d} (q – r) + { A + (q -1)d} (r –p) + { A + (r -1)d}(p – q)

= A{(q – r) + (r – p) + (p – q)} + d {(p – 1) (q – r) + (q – 1) (r –p) + (r -1)(p – q) }

= A (q – r + r – p + p – q) + d (pq – pr – q + r + qr – pq – r + p + pr – qr – p + q)

= A × 0 + d× 0

= 0 Proved

Given that t_{n} = 2n+1

Putting n = 1, 2, 3 respectively;

t_{1} = 2 x 1 + 1 = 3

t_{2} = 2 x 2 + 1 = 5

t_{3} = 2 x 3 + 1 = 7

A.P. : 3, 5, 7 .......... 2n+1 **Answer**

Given that a = 105, d = 98 - 105 = - 7

Let the n^{th }term is zero

**∴ a _{n} = a + (n - 1) d**

0 = 105 + (n - 1) - 7

0 = 105 - 7n + 7

7n = 112

n = 112/7

n = 16 **Answer**

Given that : a = 114, d = 109 - 114 = - 5

Let the first negative term n

**∴ a _{n} = - 1**

** a _{n} = a + (n - 1) d**

- 1 = 114 + (n - 1) - 5

- 1 = 114 - 5n + 5

5n = 114 + 5 + 1

5n = 120

n = 120/5

n = 24 Answer

Hence 24th term is the frist negative

**Important Question for board and preboard exams**

Given that : S_{n} = 3n^{2} + 4

Now Putting n = 1

S_{1} = 3(1)^{2} + 4

= 3 + 4

= 7

Putting n = 2

S_{2} = 3(2)^{2} + 4

= 3 x 4 + 4

= 12 + 4

= 16

a_{1} ≠** **S_{1} ≠ 7

**∵ **a_{1} = S_{1} - S_{0 } Here S_{0} ≠ 0

S_{0} = 3n^{2} + 4 **⇒ **3(0)^{2} + 4 **⇒ **0 + 4 = 4

Using a_{n} = S_{n} - S_{(n - 1)}

a_{1} = S_{1} - S_{0 }

= 7 - 4

= 3

a_{2} = S_{2} - S_{1} [ using formulla a_{n} = S_{n} - S_{(n - 1)} ]

= 16 - 7

= 9

d = a_{2} - a_{1} **⇒ **9 - 3 = 6

a_{n} = a + (n -1)d

= 3 + (n - 1) 6

= 3 + 6n - 6

= 6n - 3

Hence n^{th} term is 6n - 3 **Answer**

**Second Method : **

Given that : S_{n} = 3n^{2} + 4 ................ (i)

Replace n by (n - 1)

We have : S_{n - 1} = 3(n - 1)^{2} + 4

**⇒ ** S_{n - 1} = 3(n^{2} - 2n + 1) + 4

= 3n^{2} - 6n + 3 + 4

= 3n^{2} - 6n + 7 ......................... (ii)

a_{n} = S_{n} - S_{(n - 1) }

= 3n^{2} + 4 - (3n^{2} - 6n + 7 )

= 3n^{2} + 4 - 3n^{2} + 6n - 7

= 6n - 3 **Answer **

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