# Arithmetic Progressions Mathematics Class 10 ncert questions Answers

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## Arithmetic Progressions Mathematics Class 10 ncert questions Answers

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# Class 10 Mathematics Questions With Answers

#### Class 10 Mathematics Questions And Answers:

Gievn that a= 2k - 7, a2 = k + 5 and a3 = 3k + 2

d = a2 - a1 = k + 5 - (2k - 7) ............ (i)

d = a3 - a2 = 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

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 9th 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(am) = n(an)

m[a + (m – 1)d] = n[a + (n – 1)d]

am + m(m – 1)d = an + n(n – 1)d

am – an + [m2 – m – n+ n]d = 0

a(m – n ) + [m2 – n–( 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)

am + n = a + {(m + n)– 1}d

am + n = 0     From Equation (i)

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

ap = a =  A + (p -1)d

a = A + (p -1)d    …………. (1)

aq = b = A + (q -1)d

b = A + (q -1)d    …………. (2)

ar = 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 tn = 2n+1

Putting n = 1, 2, 3 respectively;

t1 = 2 x 1 + 1 = 3

t2 = 2 x 2 + 1 = 5

t3 = 2 x 3 + 1 = 7

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

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

Let the nth term is zero

∴ an = a + (n - 1) d

0 = 105 + (n - 1) - 7

0 = 105 - 7n + 7

7n = 112

n = 112/7

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

Let the first negative term n

∴ an =  - 1

an = a + (n - 1) d

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

- 1 = 114 - 5n + 5

5n = 114 + 5 + 1

5n = 120

n = 120/5

Hence 24th term is the frist negative

Important Question for board and preboard exams

Given that :   Sn = 3n2 + 4

Now Putting n = 1

S1 = 3(1)2 + 4

= 3 + 4

= 7

Putting n = 2

S2 = 3(2)2 + 4

= 3 x 4 + 4

= 12 + 4

= 16

a1 ≠ S1 ≠ 7

∵ a1 = S1 - S  Here S0 ≠  0

S0 = 3n2 + 4 ⇒ 3(0)2 + 4  ⇒ 0 + 4 = 4

Using  an = Sn - S(n - 1)

a1 = S1 - S

= 7 - 4

= 3

a2 = S2 - S1   [ using formulla an = Sn - S(n - 1) ]

= 16 - 7

= 9

d = a2 - a1 ⇒ 9 - 3 = 6

an = a + (n -1)d

= 3 + (n - 1) 6

= 3 + 6n - 6

= 6n - 3

Hence nth term is 6n - 3 Answer

Second Method :

Given that :   Sn = 3n2 + 4  ................ (i)

Replace n by (n - 1)

We have : Sn - 1 = 3(n - 1)2 + 4

⇒  Sn - 1 = 3(n2 - 2n + 1) + 4

=  3n2 - 6n + 3 + 4

=  3n2 - 6n + 7   ......................... (ii)

an = Sn - S(n - 1)

= 3n2 + 4 - (3n2 - 6n + 7 )

= 3n2 + 4 - 3n2 + 6n - 7

### Study Materials

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#### Class 10 Mathematics Arithmetic Progressions 2023-2024 Latest Syllabus

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