Advertisements
Advertisements
Question
The common difference of the A.P.
Options
- \[\frac{1}{3}\]
- \[- \frac{1}{3}\]
−b
b
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
= \[\frac{1 - 3b}{3} - \frac{1}{3}\]
= \[\frac{- 3b}{3} = - b\]
APPEARS IN
RELATED QUESTIONS
Find the sum of the following APs.
−37, −33, −29, …, to 12 terms.
In an AP given a = 8, an = 62, Sn = 210, find n and d.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed, and how many logs are in the top row?
If (m + 1)th term of an A.P is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term.
Find the sum of all odd numbers between 100 and 200.
The first and last terms of an AP are a and l respectively. Show that the sum of the nth term from the beginning and the nth term form the end is ( a + l ).
A sum of ₹2800 is to be used to award four prizes. If each prize after the first is ₹200 less than the preceding prize, find the value of each of the prizes
Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.
The sum of the first n terms of an AP is (3n2+6n) . Find the nth term and the 15th term of this AP.
Draw a triangle PQR in which QR = 6 cm, PQ = 5 cm and times the corresponding sides of ΔPQR?
Find the sum of the first 15 terms of each of the following sequences having nth term as xn = 6 − n .
The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by \[\frac{l^2 - a^2}{k - (l + a)}\] , then k =
The first three terms of an A.P. respectively are 3y − 1, 3y + 5 and 5y + 1. Then, y equals
If the third term of an A.P. is 1 and 6th term is – 11, find the sum of its first 32 terms.
In an A.P. a = 2 and d = 3, then find S12.
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`.
Find the sum of the integers between 100 and 200 that are
- divisible by 9
- not divisible by 9
[Hint (ii) : These numbers will be : Total numbers – Total numbers divisible by 9]
Find the middle term of the AP. 95, 86, 77, ........, – 247.
Find the sum of first 16 terms of the A.P. whose nth term is given by an = 5n – 3.
