Advertisements
Advertisements
प्रश्न
Decide whether the given sequence 24, 17, 10, 3, ...... is an A.P.? If yes find its common term (tn)
उत्तर
The given sequence is 24, 17, 10, 3, ......
Here, t1 = 24, t2 = 17, t3 = 10, t4 = 3
∴ t2 – t1 = 17 – 24 = – 7
t3 – t2 = 10 – 17 = – 7
t4 – t3 = 3 – 10 = – 7
∴ t2 – t1 = t3 – t2 = …= – 7 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P.
tn = a + (n – 1)d
= 24 + (n – 1)(– 7)
= 24 – 7n + 7
= 31 – 7n
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`
Find the sum of the following arithmetic series:
34 + 32 + 30 +...+10
Find the sum of the following arithmetic series:
(-5)+(-8)+(-11)+...+(-230)
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, ..............
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,...
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.
Choose the correct alternative answer for the following sub-question
If the third term and fifth term of an A.P. are 13 and 25 respectively, find its 7th term
Find t5 if a = 3 आणि d = −3
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, ......
12, 16, 20, 24, ...... Find 25th term of this A.P.
The nth term of an A.P. 5, 8, 11, 14, ...... is 68. Find n = ?
If p - 1, p + 3, 3p - 1 are in AP, then p is equal to ______.
