Arithmetic Progressions Mathematics Class 10 ncert questions Answers


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Arithmetic Progressions Mathematics Class 10 ncert questions Answers


All chapters of NCERT Book as ncert solutions have exercise questions, textual questions and so many addtional questions like short answered questions, long answered questions and very long questions, here we included all types of questions answers format that need for a students and other stock holders like teachers and tutors. NCERT Solutions for Class 10 Mathematics extra and important Questions Answers Arithmetic Progressions. Our NCERT Solutions is a powerfull application for cbse learning students and other learners who always search variety of questions with answers from ncert solutions or from questions bank sample papers and board exams questions.

NCERT Solutions for Class 10 Mathematics Questions with Solutions

Class 10 Mathematics Questions With Answers

Class 10 Mathematics Questions And Answers:

Answer:

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

k = 5 Answer

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 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 

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)

Answer:

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            

Answer:

 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

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

n = 16 Answer

Answer:

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 

n = 24 Answer 

Hence 24th term is the frist negative

Important Question for board and preboard exams

Answer:

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     

     = 6n - 3    Answer 

Study Materials

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Class 10 Mathematics Arithmetic Progressions 2022-2023 Latest Syllabus

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