Consider the sequence defined by tn = an2 + bn + c. If t2 = 3, t4 = 13 and t7 = 113, show that 3tn = 17n2 – 87n + 115
Here, ...(i)
Now,
4a + 2b + c = 3 ...(ii)
16a + 4b + c = 13 ...(iii)
49a + 7b + c = 113 ...(iv)
Subtracting (iii) from (iv), we get 33a + 3b = 100 ...(v)
Subtracting (ii) from (iii), we get 12a + 2b = 10 6a + b = 5 ...(vi)
Multiplying both sides of (vi) by 3, we get
18a + 3b = 15 ...(vii)
Subtracting (vii) from (v), we get
Using this value of a in (vi), we get
Using values of a and b in (ii), we get
Using values of a, b, c in (i), we get
Find the indicated terms in the following sequence whose nth terms are :
Here,
Putting n = 19, we get
Show that the sequence defined by
(where A and B are constant) is an A.P. with common difference A.
Here,
Replacing n by n - 1, we get
which is constant and independent of n.
Hence, the sequence is an A.P.
Find the indicated terms in the following sequence whose nth terms are:
Here, ...(i)
Replacing n by n-1 in (1), we get
Putting n = 16 in (I), we get
Find the first six terms of the sequence whose first term is 1 and whose (n+l)th term is obtained by adding n to the nth term.
Let denote the nth term
According to given equation ...(i)
Putting n = 1 in (i), we get
Putting n = 2 in (i), we get
Putting n = 3 in (i), we get
Putting n = 4 in (i), we get
Putting n = 5 in (i), we get