English

The common difference of the A.P. 1 2 b , 1 − 6 b 2 b , 1 − 12 b 2 b , . . . is - Mathematics

Advertisements
Advertisements

Question

The common difference of the A.P. \[\frac{1}{2b}, \frac{1 - 6b}{2b}, \frac{1 - 12b}{2b}, . . .\] is 

 

Options

  •  2b

  • −2b

  • 3

  •  - 3 

MCQ
Advertisements

Solution

Let a be the first term and d be the common difference.
The given A.P. is  \[\frac{1}{2b}, \frac{1 - 6b}{2b}, \frac{1 - 12b}{2b}, . . .\]

Common difference = d = Second term − First term
                                       = \[\frac{1 - 6b}{2b} - \frac{1}{2b}\]

                                       = \[\frac{- 6b}{2b} = - 3\]

 

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Arithmetic Progression - Exercise 5.8 [Page 59]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 5 Arithmetic Progression
Exercise 5.8 | Q 39 | Page 59

RELATED QUESTIONS

The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of x such that sum of numbers of houses preceding the house numbered x is equal to sum of the numbers of houses following x.


Find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 3


Find the sum of the following arithmetic progressions:

3, 9/2, 6, 15/2, ... to 25 terms


How many two-digit number are divisible by 6?


Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP.


In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms. 


In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
(Assume that three consecutive terms in A.P. are a – d, a, a + d).


In an A.P. the 10th term is 46 sum of the 5th and 7th term is 52. Find the A.P.


Find the A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.


Find the sum  (−5) + (−8)+ (−11) + ... + (−230) .


The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.


The common difference of an A.P., the sum of whose n terms is Sn, is


Let the four terms of the AP be a − 3da − da + and a + 3d. find A.P.


 Q.10


Q.15


Q.17 


Find the sum of natural numbers between 1 to 140, which are divisible by 4.

Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136

Here d = 4, therefore this sequence is an A.P.

a = 4, d = 4, tn = 136, Sn = ?

tn = a + (n – 1)d

`square` = 4 + (n – 1) × 4

`square` = (n – 1) × 4

n = `square`

Now,

Sn = `"n"/2["a" + "t"_"n"]`

Sn = 17 × `square`

Sn = `square`

Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.


How many terms of the AP: –15, –13, –11,... are needed to make the sum –55? Explain the reason for double answer.


Find the sum of all even numbers from 1 to 250.


Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60...... is 300?

Hence find the sum of all the terms of the Arithmetic Progression (A.P.)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×