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A sequence of numbers <a_{n}> is said to be in arithmetic progression if the difference between any two terms of the sequence is constant. The constant difference between the two terms is called the common difference.
If ‘a’ is the first term of the sequence and ‘d’ is the common difference then the nth term of A.P. is given by
t_{n} = a + (n-1)d
In this case, the sum of first n terms of the A.P. is given by:
S_{n} = n/2 [2a + (n-1)d]
Watch this Video for more reference
We discuss some of the problems based on arithmetic progression:
Illustration: Let sum of n terms of a series be n (2n–1). Find its m^{th} term.
Solution: Let S_{m} and S_{m–1} denote the sum of first m and (m – 1) terms respectively.
S_{m} = T_{1} + T_{2} + T_{3} + ……. + T_{m–1} + T_{m}
S_{m–1} = T_{1} + T_{2} + T_{3} + ……. + T_{m–1}
Subtracting the two equations
S_{m} – S_{m–1} = T_{m}
⇒ T_{m} = (m(2m–1))–(m–1)(2(m–1)–1))
= (2m^{2} – m)–(2m^{2} – 5m + 3)
= 4m – 3
Illustration: The sum of n terms of two A.P.’s is in the ratio 3n+2: 2n+3. Find the ratio of their 10^{th} terms.
Solution: let us assume the two A.P’s to be of the forms
Let a, a + d, a + 2d, a + 3d, ……………
A, A + D, A + 2D, A + 3D, ………………
It is given in the question that
⇒ To get the ratio of 10^{th} terms put n–1/2 = 9
or n = 19
⇒ a+9d/A+9D = 3(19)+2/2(19)+3 = 59/41
Illustration: Let T_{r} be the r^{th} term of an A.P. for r =1, 2, 3, ….. if for some positive integers m and n we have T_{m} = 1/n and T_{n} = 1/m, then find the value of T_{mn}.
Solution: Let T_{m} = a + (m-1)d = 1/n
T_{n} = a + (n-1)d = 1/m
On subtracting the second equation from the first equation, we get,
(m-n)d = 1/n – 1/m =(m-n)/mn
Hence, this gives d = 1/mn
Again, T_{mn} = a + (mn-1)d
= a+(mn – n + n -1)d
= a+(n-1)d+(mn-n)d
= T_{n }+n(m-1)1/mn
= 1/m + (m-1)/m =
Hence T_{mn} = 1.
Illustration: The fourth power of the common difference of an A.P. with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.
Solution: Let four consecutive terms of the A.P. a-3d, a-d, a+d, a+3d
Then the product is given by
P = (a-3d)(a-d)(a+d)(a+3d) + (2d)^{4}
= (a^{2}-9d^{2}) (a^{2}-d^{2}) + 16d^{4}
= (a^{2}-5d^{2})^{2}
Now, (a^{2}-5d^{2}) = a^{2}-9d^{2} + 4d^{2}
(a-3d)(a+3d) + (2d)^{2}
= I.I + I^{2}
= I
Therefore, P = I^{2 }= Integer.
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