IIT JEE Sum of First N Terms Of AP | JEE The Sum of First N Terms of AN AP

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]
 

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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 = T₁ + T₂ + T₃ + ……. + T_m–1 + T_m

S_m–1 = T₁ + T₂ + T₃ + ……. + 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))

         = (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 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

   = (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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