Sum of first n terms of AP

A sequence of numbers <an> 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

tn = a + (n-1)d

In this case, the sum of first n terms of the A.P. is given by:

Sn = n/2 [2a + (n-1)d]

View the video on arithmetic progression

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 mth term.

Solution: Let Sm and Sm–1 denote the sum of first m and (m – 1) terms respectively.

Sm = T1 + T2 + T3 + ……. + Tm–1 + Tm

Sm–1 = T1 + T2 + T3 + ……. + Tm–1

Subtracting the two equations

Sm – Sm–1 = Tm

⇒ Tm = (m(2m–1))–(m–1)(2(m–1)–1))

         = (2m2 – m)–(2m2 – 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 10th 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 10th terms put n–1/2 = 9

or n = 19

⇒ a+9d/A+9D = 3(19)+2/2(19)+3 = 59/41

Illustration: Let Tr be the rth term of an A.P. for r =1, 2, 3, ….. if for some positive integers m and n we have Tm = 1/n and Tn = 1/m, then find the value of Tmn.

Solution: Let Tm = a + (m-1)d = 1/n

Tn = 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, Tmn = a + (mn-1)d

                 = a+(mn – n + n -1)d

                 = a+(n-1)d+(mn-n)d

                 = Tn +n(m-1)1/mn

                 = 1/m + (m-1)/m =

Hence Tmn = 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

   = (a2-9d2) (a2-d2) + 16d4

   = (a2-5d2)2

Now, (a2-5d2) = a2-9d2 + 4d2

(a-3d)(a+3d) + (2d)2

= I.I + I2

= I

Therefore, P = I2 = Integer.

 

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