# Important extra questions for Class 10 maths chapter 5. Arithmetic Progressions - Questions Bank

Important extra questions for class 10 maths chapter 5 arithmetic progressions. We hope this blog will help the learners as well as those candicate who will appear in the boards exams 2023-2024 because we provide here all possible solutions of queries by our users. Important extra questions for Class 10 maths chapter 5. Arithmetic Progressions - Questions Bank (Important extra questions for class 10 maths chapter 5 arithmetic progressions) for Examination Year 2023-2024

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

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

Arithmetic Progressions | Samantar Shredhi | समांतर श्रेढ़ी

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

k के किस मान के लिए 2k-7,k+5 और 3k+2  समान्तर श्रेणी के क्रमागत पद होंगे?

Solution:

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

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 का मान ज्ञात कीजिए जिसके लिए (8x + 4), (6x – 2)  और (2x + 7) समांतर श्रेढ़ी में है।

Solution: इसका भी हल प्रश्न Question 1. की तरह ही होगा, विद्यार्थी स्वयं करें |

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.

दो समांतर श्रेढियों के n वें पदों के योग का अनुपात 3n + 8 : 7n + 15 है तो उनके 9 वें पदों का अनुपात ज्ञात कीजिये |

Solution:

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

यदि किसी A.P का m वाँ पद का m गुणा उसी A.P के n वें पद के n गुणा के बराबर है तो दर्शाइए कि (m + n) वाँ पद शून्य है |

Or

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

यदि किसी A.P का m वाँ पद का m गुणा उसी A.P के n वें पद के n गुणा के बराबर है तो बताइए कि कौन सा पद शून्य होगा ?

Solution: माना कि समान्तर श्रेढ़ी का प्रथम पद a तथा सार्व अंतर d है |

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 – 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)

am + n = a + {(m + n)– 1}d

am + n = 0     समीकरण (1) से

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

यदि एक समान्तर श्रेढ़ी का p वाँ, q वाँ तथा r वाँ पद क्रमश: a, b तथा c हों, तो सिद्ध कीजिए कि 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.

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

Q6. चार संख्याएँ समांतर श्रेढ़ी में हैं | यदि उनका योगफल 20 है और तथा उनके वर्गों का योग 120 हो तो संख्याएँ ज्ञात कीजिए |

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

यदि किसी A.P. का tn = 2n+1 हैं तो A.P. ज्ञात कीजिए |

Solution:

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

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

A.P: 105, 98, 91, ........................ का कौन-सा पद शून्य होगा ?

Solution:

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

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

अनुक्रम 114, 109, 104………… का कौन-सा पद पहला ऋणात्मक पद है?

Solution:

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

Q10. If the sum of nth term of an A.P. is define as Sn = 3n2 + 4 then find the nth term?

यदि किसी समांतर श्रेणी के nवें पद का योग Sn = 3n2 + 4 के रूप में परिभाषित किया जाता है, तो nवाँ पद ज्ञात कीजिए?

Solution:

First Method: Alternative Method

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

#### Other Extra Questions for Practice on Chapter - 5  Arithmetic Progressions | Samantar Shredhi | समांतर श्रेढ़ी

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 nth term of an A.P  4, 7, 10, ..................... is 82. Then find the value of n.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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, ......................

## Chapter: Important extra questions for Class 10 maths chapter 5. Arithmetic Progressions | Topic: Arithmetic Progressions | Samantar Shredhi | समांतर श्रेढ़ी

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# Important extra questions for Class 10 maths chapter 5. Arithmetic Progressions - Questions Bank

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. Important extra questions for Class 10 maths chapter 5. Arithmetic Progressions - Questions Bank (Important extra questions for class 10 maths chapter 5 arithmetic progressions) for Examination Year 2023-2024