English

The Common Difference of the A.P. 1 3 , 1 − 3 B 3 , 1 − 6 B 3 , . . . is - Mathematics

Advertisements
Advertisements

Question

The common difference of the A.P.

\[\frac{1}{3}, \frac{1 - 3b}{3}, \frac{1 - 6b}{3}, . . .\] is 
 

Options

  • \[\frac{1}{3}\]

     

  • \[- \frac{1}{3}\]

     

  • b

  • b

MCQ
Advertisements

Solution

Let a be the first term and d be the common difference.
The given A.P. is   \[\frac{1}{3}, \frac{1 - 3b}{3}, \frac{1 - 6b}{3}, . . .\]

Common difference = d = Second term − First term
                                       = \[\frac{1 - 3b}{3} - \frac{1}{3}\]

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

 

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 38 | Page 59

RELATED QUESTIONS

If the 12th term of an A.P. is −13 and the sum of the first four terms is 24, what is the sum of first 10 terms?


Find the sum of the first 51 terms of the A.P: whose second term is 2 and the fourth term is 8.


Find four numbers in AP whose sum is 8 and the sum of whose squares is 216.


Find the sum of first n even natural numbers.


Find the sum of first n terms of an AP whose nth term is (5 - 6n). Hence, find the sum of its first 20 terms.


Find the first term and common difference for  the A.P.

0.6, 0.9, 1.2,1.5,...


Sum of 1 to n natural numbers is 36, then find the value of n.


Find out the sum of all natural numbers between 1 and 145 which are divisible by 4.


For what value of p are 2p + 1, 13, 5p − 3 are three consecutive terms of an A.P.?

 

Write the nth term of the \[A . P . \frac{1}{m}, \frac{1 + m}{m}, \frac{1 + 2m}{m}, . . . .\]

 

If k, 2k − 1 and 2k + 1 are three consecutive terms of an A.P., the value of k is


Q.6


Find the sum of first 10 terms of the A.P.

4 + 6 + 8 + .............


How many terms of the series 18 + 15 + 12 + ........ when added together will give 45?


Find the sum of first 1000 positive integers.

Activity :- Let 1 + 2 + 3 + ........ + 1000

Using formula for the sum of first n terms of an A.P.,

Sn = `square`

S1000 = `square/2 (1 + 1000)`

= 500 × 1001

= `square`

Therefore, Sum of the first 1000 positive integer is `square`


Shubhankar invested in a national savings certificate scheme. In the first year he invested ₹ 500, in the second year ₹ 700, in the third year ₹ 900 and so on. Find the total amount that he invested in 12 years


Find the sum:

`(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) +` ... to 11 terms


If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.


The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.


The sum of 40 terms of the A.P. 7 + 10 + 13 + 16 + .......... is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×