Answer:

Important extra questions for Class 10 maths chapter 5. Arithmetic Progressions

Arithmetic Progressions | Samantar Shredhi | ?????? ??????

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Q1. For what value of k, 2k-7,k+5 and 3k+2 are the consecutive terms of an A.P?

       k ?? ??? ??? ?? ??? 2k-7,k+5 ?? 3k+2  ??????? ?????? ?? ??????? ?? ??????

Solution: 

Gievn that a₁ = 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

Similar Extra and Important question you can try this;

Q2. Find the value of x for which (8x + 4), (6x – 2)  and (2x + 7) are in sequence.

      x ?? ??? ????? ????? ????? ??? (8x + 4), (6x – 2)  ?? (2x + 7) ?????? ?????? ??? ??? 

Solution: ???? ?? ?? ?????? Question 1. ?? ??? ?? ????, ?????????? ????? ???? | 

Answer = 7

Q3. The sum of n terms of two A.P.’s is in the ratio 3n + 8: 7n + 15. Find the ratio of their 9th terms.

         ?? ?????? ????????? ?? n ??? ???? ?? ??? ?? ?????? 3n + 8 : 7n + 15 ?? ?? ???? 9 ??? ???? ?? ?????? ????? ?????? |

Solution: 

Q4. If m times of m^th term of an A.P. is equal to n times of n^th term of same A.P. Show that (m + n)^th term is zero.

??? ???? A.P ?? m ??? ?? ?? m ???? ??? A.P ?? n ??? ?? ?? n ???? ?? ????? ?? ?? ??????? ?? (m + n) ??? ?? ????? ?? |

Or 

If m times of m^th term of an A.P. is equal to n times of n^th term of same A.P. Which term is zero?

??? ???? A.P ?? m ??? ?? ?? m ???? ??? A.P ?? n ??? ?? ?? n ???? ?? ????? ?? ?? ????? ?? ??? ?? ?? ????? ???? ?

Solution: ???? ?? ??????? ?????? ?? ????? ?? a ??? ????? ???? d ?? |

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

??: 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 + [m² – m – n² + n]d = 0

⇒ a(m – n ) + [m² – n² –( 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     ?????? (1) ??

Q5. if p^th , q^th and r^th terms of an A.P. is a, b and c respectively, prove that : a(q – r) + b(r – p) + c(p –q) = 0 .

??? ?? ??????? ?????? ?? p ???, q ??? ??? r ??? ?? ?????: a, b ??? c ???, ?? ????? ????? ?? a(q – r) + b(r – p) + c(p –q) = 0 .

Solutions: 

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            

Q6. ??? ???????? ?????? ?????? ??? ??? | ??? ???? ????? 20 ?? ?? ??? ???? ?????? ?? ??? 120 ?? ?? ???????? ????? ????? |

Q7. If t_n = 2n+1 then find the series (A.P).

      ??? ???? A.P. ?? t_n = 2n+1 ??? ?? A.P. ????? ????? | 

Solution: 

 Given that t_n = 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

Q8. Which term will be zero of A.P: 105, 98, 91, ........................ ?

       A.P: 105, 98, 91, ........................ ?? ???-?? ?? ????? ???? ? 

Solution: 

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

Q9. Which term of the sequence 114, 109, 104…………is the first negative term?

      ??????? 114, 109, 104………… ?? ???-?? ?? ???? ??????? ?? ???

Solution: 

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

Q10. If the sum of n^th term of an A.P. is define as S_n = 3n² + 4 then find the n^th term?

??? ???? ?????? ?????? ?? n??? ?? ?? ??? S_n = 3n² + 4 ?? ??? ??? ???????? ???? ???? ??, ?? n??? ?? ????? ??????

Solution:

First Method: Alternative Method 

Given that :   S_n = 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 - S₀   Here S₀ ≠  0

S₀ = 3n² + 4 ⇒ 3(0)² + 4  ⇒ 0 + 4 = 4 

Using  a_n = S_n - S_{(n - 1)} 

a₁ = S₁ - S₀ 

     = 7 - 4 

     = 3 

a₂ = S₂ - S₁   [ using formulla a_n = S_n - S_{(n - 1)} ]

     = 16 - 7 

     = 9 

d = a₂ - a₁ ⇒ 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 = 3n² + 4  ................ (i) 

Replace n by (n - 1) 

We have : S_{n - 1} = 3(n - 1)² + 4  

⇒  S_{n - 1} = 3(n² - 2n + 1) + 4  

               =  3n² - 6n + 3 + 4 

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

a_n = S_n - S_{(n - 1)} 

     = 3n² + 4 - (3n² - 6n + 7 )    

     = 3n² + 4 - 3n² + 6n - 7     

     = 6n - 3    Answer 

Other Extra Questions for Practice on Chapter - 5  Arithmetic Progressions | Samantar Shredhi | ?????? ??????

Q1.  How many three-digit numbers are divisible by 7?

Q2.  How many multiples of 5 are between 1 to 102?  

Q3.  How many multiples of 4 are between 10 to 250?                                                                                            

Q3. Find the A.P whose third term is -13 and sixth term is 2. 

Answer : –23, –18, –13, .......

Q4.  if n^th term of an A.P  4, 7, 10, ..................... is 82. Then find the value of n. 

Q5. if n^th term of an A.P is 2n + 7, then find the 7th term of A.P. 

Q6. if n^th term of an A.P is 2n - 8, what term is zero of the A.P? 

Q7. Find 25^th term of an A.P if a_n = 3n - 2. 

Q8. The final term of an A.P: 9, 17, 25, ................... is 601, find 35^th term from final term. 

Q9.  Which term will be zero of A.P: 105, 98, 91, ........................

Q10.  Find 41^st term of an A.P if a_n = 3 - 4n.

Q11.  FInd the numbers of all multiples of 4 between 19 to 101. 

Q12. if n^th term of an A.P is 9 - 5n, then find the 21^st term. 

Q13.  Find the 21^st term of  A.P 9, 11, 13, 15, .........................                                                                                       Answer : 49   

Q14.  Which term is 78 of A.P 3, 8, 13, 18, ..........................

Answer : 16th term 

Q15.  If 12, x and 22 are in A.P then find the value of x. 

 Answer : 17 

Q16.  Find the 12^th term of A.P 3, 9, 15 .................. 

Answer: 69 

Q17.  Find the expression for n^th term of A.P: - 5, -2, 1, ..........................

 Answer: 3n - 8 

Q18.  if three numbers 5, 2k - 3 and 9 are in A.P then find the value of x. 

 Answer : 5 

Q19.  Find the 9^th term from last of the A. P: 7, 11, 15, .................... 147

  Answer : 115 

Q20.  Which term of A.P: 41, 38, 35, ...................... is the first negative? 

Answer : 15th term is negative 

Q21.  Find first negative term of A.P:  41, 38, 35, ......................