Advertisements
Advertisements
Question
If the first term of an A.P. is a and nth term is b, then its common difference is
Options
- \[\frac{b - a}{n + 1}\]
- \[\frac{b - a}{n - 1}\]
- \[\frac{b - a}{n}\]
- \[\frac{b + a}{n - 1}\]
Solution
Here, we are given the first term of the A.P. as a and the nth term (an) as b. So, let us take the common difference of the A.P. as d.
Now, as we know,
an = a + ( n-1) d
On substituting the values given in the question, we get.
b = a + ( n - 1) d
( n - 1) d = b - a
d = \[\frac{b - a}{n - 1}\]
Therefore, d = \[\frac{b - a}{n - 1}\]
APPEARS IN
RELATED QUESTIONS
How many multiples of 4 lie between 10 and 250?
In an AP Given a12 = 37, d = 3, find a and S12.
Find the sum to n term of the A.P. 5, 2, −1, −4, −7, ...,
The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.
Find the sum of all natural numbers between 200 and 400 which are divisible by 7.
If the seventh term of an A.P. is \[\frac{1}{9}\] and its ninth term is \[\frac{1}{7}\] , find its (63)rd term.
The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25th term of this A.P.
The sum of the first three terms of an Arithmetic Progression (A.P.) is 42 and the product of the first and third term is 52. Find the first term and the common difference.
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`.
A merchant borrows ₹ 1000 and agrees to repay its interest ₹ 140 with principal in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 10. Find the amount of the first instalment
