If for a sequence (tn),sn=2n²+5n,find tn and show that the sequence is an a.p
s(1) = t(1) = 2+5 = 7
s(2) = 2(2^2) + 5(2) = 18
s(3) = 2(9) + 5(3) = 33
s(4) = 2(16) + 5(4) = 52
..
t(2) = s(2) - s(1) = 18-7 = 11
t(3) = s(3) - s(2) = 33-18
= 15
t(4) = s(4) - s(3) = 52-33 = 19
so we have:
7,11,15,19,... looks like an AP with
a = 7 , d = 4
term(n) = a + (n-1)d
= 7 + 4(n-1)
= 7 + 4n - 4
= 4n + 3
in general:
term(n) = sum(n) - sum(n-1)
= 2n^2 + 5n - (2(n-1)^2 + 5(n-1) )
= 2n^2 + 5n - ( 2n^2 - 4n + 2 + 5n - 5)
= 2n^2 + 5n - 2n^2 -n + 3
= 4n + 3
To find the nth term of a sequence, tn, we need to understand the pattern of the sequence.
Given: sn = 2n² + 5n
We can express the nth term of a sequence, tn, in terms of sn.
Let's find tn:
To find tn, we substitute the n value into the given expression for sn.
tn = 2n² + 5n
Now let's simplify it further:
tn = 2n(n + 5)
Now, let's prove that the sequence is an arithmetic progression (a.p).
To prove that the sequence is an a.p., we need to show that the difference between any two consecutive terms is constant.
Let's find the difference between the (n+1)th term and the nth term:
tn+1 = 2(n+1)² + 5(n+1)
tn+1 = 2(n² + 2n + 1) + 5n + 5
tn+1 = 2n² + 4n + 2 + 5n + 5
tn+1 = 2n² + 5n + 4n + 2 + 5
tn+1 = (2n² + 5n) + (4n + 7)
Now let's find the difference between tn+1 and tn:
tn+1 - tn = (2n² + 5n) + (4n + 7) - (2n² + 5n)
tn+1 - tn = 4n + 7
The difference is 4n + 7, which is not a constant value. Hence, the given sequence (sn) is not an arithmetic progression (a.p).
However, the formula for tn, tn = 2n(n + 5), is correct.
To find tn for the given sequence (sn = 2n² + 5n), we need to substitute the value of n into the given formula.
The general formula for an arithmetic progression (a.p.) can be written as:
tn = a + (n-1)d
Here, we have sn = 2n² + 5n, which represents the sum of the first n terms. Now, let's find tn and show that the sequence is an arithmetic progression.
Step 1: Find tn
To find tn, substitute n into the expression sn = 2n² + 5n:
tn = 2n² + 5n
So, the formula for tn of the given sequence is tn = 2n² + 5n.
Step 2: Show that the sequence is an arithmetic progression
For a sequence to be an arithmetic progression, the difference between any two consecutive terms must be constant.
To verify if the given sequence is an arithmetic progression, let's calculate the difference between two consecutive terms:
tn+1 - tn
Substituting the formula tn = 2n² + 5n, we get:
tn+1 = 2(n+1)² + 5(n+1)
tn+1 = 2(n² + 2n + 1) + 5n + 5
tn+1 = 2n² + 4n + 2 + 5n + 5
tn+1 = 2n² + 9n + 7
Now, calculate the difference:
tn+1 - tn = (2n² + 9n + 7) - (2n² + 5n)
tn+1 - tn = 2n² + 9n + 7 - 2n² - 5n
tn+1 - tn = (2n² - 2n²) + (9n - 5n) + 7
tn+1 - tn = 4n + 7
Since the difference between consecutive terms (tn+1 - tn) is a constant (4n + 7), we can conclude that the given sequence is an arithmetic progression.