Find the common difference of the AP . (2013D)

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Find the common difference of the A.P. . (2013D)

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d = a_{2} – a_{1} =

= = -3

Answer :

d = a_{2} – a_{1} =

= = -2

Calculate the common difference of the A.P. . (2013D)

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d = a_{2} – a_{1}=

=

Calculate the common difference of the A.P. (2013OD)

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d = a_{2} – a_{1} =

= = -b

What is the common difference of an A.P. in which a_{21} – a_{7} = 84? (2017OD )

Answer :

a_{21} – a_{7} = 84 …[Given in the question

∴ (a + 20d) – (a + 6d) = 84 …[a_{n} = a + (n – 1)d

20d – 6d = 84

14d = 84 ⇒ d = 6

Find the 9th term from the end (towards the first term) of the A.P. 5,9,13, …, 185. (2016D)

Answer :

Here, a = 5, d = 9 – 5 = 4, l = 185

9^{th} term from the end = 185 – (9 – 1)4

= 185 – 8 × 4 = 185 – 32 = 153

The angles of a triangle are in A.P., the least being half the greatest. Find the angles. (2011D)

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40°, 60°, 80°

Is -150 a term of the A.P. 17, 12, 7, 2, … ? (2011D)

Answer :

no

Which term of the progression 4, 9, 14, 19, … is 109? (2011D)

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22nd term

If p^{th}, q^{th} and r^{th} terms of an A.P. are a, b, c respectively, then show that (a – b)r + (b – c)p+ (c – a)q = 0. (2011D)

Answer :

Let A be the first term and D be the common difference of the given A.P.

p^{th} term = A + (p – 1)D = a …(A)

q^{th} term = A + (q – 1)D = b …(B)

r^{th} term = A + (r – 1)D = c … (C)

L.H.S. = (a – b)r + (b – c)p + (c – a)q

= [A + (p – 1)D – (A + (q – 1)D)]r + [A + (q – 1)D – (A + (r – 1)D)]p + [A + (r – 1)D – (A + (p – 1)D)]q

= [(p – 1 – q + 1)D]r + [(q – 1 – r + 1)D]p + [(r – 1 – p + 1)D]q

= D[(p – q)r + (q – r)p + (r – p)q]

= D[pr – qr + qp – rp + rq – pq]

= D[0] = 0 = R.H.S.

The 17^{th} term of an AP is 5 more than twice its 8^{th} term. If the 11^{th} term of the AP is 43, then find its n^{th} term. (2012D)

Answer :

From (i),

a = 2(4) – 5 = 8 – 5 = 3

As a_{n} = a + (n – 1) d

∴ a_{n} = 3 + (n – 1) 4 = 3 + 4n – 4

a_{n} = (4n – 1)

The 15^{th} term of an AP is 3 more than twice its 7^{th} term. If the 10^{th} term of the AP is 41, then find its n^{th} term. (2012D)

Answer :

From (i),

a = 2(4) – 3

= 8 – 3 = 5

n^{th} term = a + (n – 1) d

∴ n^{th} term = 5 + (n – 1) 4

= 5 + 4n – 4 = (4n + 1)

The 16^{th} term of an AP is 1 more than twice its 8^{th} term. If the 12^{th} term of the AP is 47, then find its n^{th} term. (2012D)

Answer :

From (A) and (B), a = 4 – 1 = 3

As n^{th} term = a + (n – 1) d

∴ n^{th} term = 3 + (n – 1) 4

= 3 + 4n -4 = 4n -1

A sum of ₹1,600 is to be used to give ten cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes. (2012OD)

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Here, Sum of 10 prizes = 1600

S_{10} = 1600, d = -20, n = 10

S_{n} = (2a + (n – 1)d]

∴ [2a + (10 – 1)(-20)] = 1600

2a – 180 = 320

2a = 320 + 180 = 500

a = 250

Hence, 1^{st} prize = a = ₹250

2^{nd} prize = a_{2} = a + d = 250 + (-20) = ₹230

3^{rd} prize = a_{3} = a_{2} + d = 230 – 20 = ₹210

4^{th} prize = a_{4} = a_{3} + d = 210 – 20 = ₹190

5^{th} prize = a_{5} = a_{4} + d = 190 – 20 = ₹170

6^{th} prize = a_{6} = a_{5} + d = 170 – 20 = ₹150

7^{th} prize = a_{7} = a_{6} + d = 150 – 20 = ₹130

8^{th} prize = a_{8} = a_{7} + d = 130 – 20 = ₹110

9^{th} prize = a_{9} = a_{8} + d = 110 – 20 = ₹590

10^{th} prize = a_{10} = a_{9} + d = 90 – 20 = ₹70

= ₹ 1,600

Find the 10^{th }term

1 |
1,3,5,7…. |
A |
57 |

2 |
2,4,6,8…. |
B |
51 |

3 |
2,7,12,17…. |
C |
19 |

4 |
21,24,27,30…. |
D |
20 |

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Answer :

1-C, 2-D, 3-A, 4-B

Find the n^{th }term

1 |
2,6,10,14…. |
A |
8+10n |

2 |
13,9,5,1…. |
B |
4n-2 |

3 |
18,28,38,48…. |
C |
26-n |

4 |
25,24,23,22,…. |
D |
17-4n |

Answer :

1-B, 2-D, 3-A, 4-C

Find the 2^{nd }term from last

1 |
-2,5,…….,68 |
A |
900 |

2 |
-100,0,….., 1000 |
B |
98 |

3 |
28,30,…… , 100 |
C |
15 |

4 |
75,45,….-15 |
D |
61 |

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Answer :

1-D, 2-A, 3-B, 4-C

Answer :

1-B, 2-A, 3-D, 4-C

Find the common differences

1 |
89,85,81,77 |
A |
11 |

2 |
100, 89, 78, 67 |
B |
-4 |

3 |
100, 111,122,133 |
C |
4 |

4 |
89, 93, 97, 101 |
D |
-11 |

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Answer :

1-B, 2-D, 3-A, 4-C

In this chapter, students will discuss patterns in succeeding terms obtained by adding a fixed number to the preceding terms. They also, see how to find nth terms and the sum of n consecutive terms. Students will learn arithmetic progression effectively when they solve daily life problems.

This chapter has Arithmetic Progression Derivation of the nth term and sum of the first n terms of an A.P. and their application in solving daily life problems. This is one of the important chapters from the point of the Class 10 examination. An arithmetic progression is a very basic and important topic to study as almost all the competitive exams will ask questions on arithmetic progression.

- Science Book for CBSE Class 10
- Maths Book for CBSE Class 10
- English Book for CBSE Class 10
- Social Science Book for CBSE Class 10
- Hindi Book for CBSE Class 10
- Gujarati Book for CBSE Class 10

The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.