Advertisements
Advertisements
प्रश्न
The nth term of an A.P. 5, 8, 11, 14, ...... is 68. Find n = ?
उत्तर
The given A.P. is 5, 8, 11, 14, ......
Here, a = 5, d = 8 – 5 = 3, tn = 68
Since tn = a + (n – 1)d,
68 = 5 + (n – 1)(3)
∴ 68 = 5 + 3n – 3
∴ 68 = 2 + 3n
∴ 66 = 3n
∴ n = `66/3`
∴ n = 22
APPEARS IN
संबंधित प्रश्न
Find the term t15 of an A.P. : 4, 9, 14, …………..
Find the sum of the following arithmetic series:
`7 + 10 1/2 + 14 + ....... + 84`
Decide whether the following sequence is an A.P., if so find the 20th term of the progression:
–12, –5, 2, 9, 16, 23, 30, ..............
Given Arithmetic Progression 12, 16, 20, 24, . . . Find the 24th term of this progression.
Find the 19th term of the following A.P.:
7, 13, 19, 25, ...
Find the 27th term of the following A.P.
9, 4, –1, –6, –11,...
Six year before, the age of mother was equal to the square of her son's age. Three year hence, her age will be thrice the age of her son then. Find the present ages of the mother and son.
Select the correct alternative and write it.
What is the sum of first n natural numbers ?
Select the correct alternative and write it.
If a share is at premium, then -
Find the 23rd term of the following A.P.: 9, 4,-1,-6,-11.
How many multiples of 4 lie between 10 and 205?
Find t5 if a = 3 आणि d = −3
Decide whether the given sequence 24, 17, 10, 3, ...... is an A.P.? If yes find its common term (tn)
How many two-digit numbers are divisible by 5?
Activity :- Two-digit numbers divisible by 5 are, 10, 15, 20, ......, 95.
Here, d = 5, therefore this sequence is an A.P.
Here, a = 10, d = 5, tn = 95, n = ?
tn = a + (n − 1) `square`
`square` = 10 + (n – 1) × 5
`square` = (n – 1) × 5
`square` = (n – 1)
Therefore n = `square`
There are `square` two-digit numbers divisible by 5
Decide whether 301 is term of given sequence 5, 11, 17, 23, .....
Activity :- Here, d = `square`, therefore this sequence is an A.P.
a = 5, d = `square`
Let nth term of this A.P. be 301
tn = a + (n – 1) `square`
301 = 5 + (n – 1) × `square`
301 = 6n – 1
n = `302/6` = `square`
But n is not positive integer.
Therefore, 301 is `square` the term of sequence 5, 11, 17, 23, ......
If tn = 2n – 5 is the nth term of an A.P., then find its first five terms
If p - 1, p + 3, 3p - 1 are in AP, then p is equal to ______.
