Advertisements
Advertisements
प्रश्न
The nth term of a sequence is 8 – 5n. Show that the sequence is an A.P.
उत्तर
tn = 8 – 5n
Replacing n by (n + 1), we get
tn+1 = 8 – 5(n + 1)
= 8 – 5n – 5
= 3 – 5n
Now,
tn+1 – tn = (3 – 5n) – (8 – 5n) = –5
Since, (tn+1 – tn) is independent of n and is therefore a constant.
Hence, the given sequence is an A.P.
APPEARS IN
संबंधित प्रश्न
Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts in 4623.
Insert four A.M.s between 14 and -1.
Q.7
Find the general term (nth term) and 23rd term of the sequence 3, 1, –1, –3, ........... .
Is –150 a term of 11, 8, 5, 2, .......?
Which term of the A.P. 3, 8, 13, 18, … is 78?
The sum of 2nd and 7th terms of an A.P. is 30. If its 15th term is 1 less than twice its 8th term, find the A.P.
If the numbers n – 2, 4n – 1 and 5n + 2 are in A.P., find the value of n.
The sum of three numbers in A.P. is 3 and their product is – 35. Find the numbers.
The sum of the first three terms of an A.P.is 33. If the product of the first and the third terms exceeds the second term by 29, find the A.P.