Sum of first n terms of APA 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)dIn this case, the sum of first n terms of the A.P. is given by:

S

_{n}= 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 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 formsLet 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 = 9or 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/nT

_{n}= a + (n-1)d = 1/mOn 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+3dThen 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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Related resources:

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