SUM OF N TERMS OF AN ARITHMETIC PROGRESSION

Sum of n Terms of an Arithmetic Progression :

To find the sum of the series, we use the formula given below.

Sn  =  (n/2) [a + l] (or)

Sn  =  (n/2) [2a + (n - 1)d]

To find the first term, common difference and number of term, we use the formula given below.

an  =  a + (n - 1)d

Sum of n Terms of an Arithmetic Progression - Examples

Question 1 :

If the sum of 7 terms of an AP is 49 and that of 17 terms is 289,find the sum of first n terms.

Solution :

S7  =  49

S17 =  289

Sn  =   (n/2) [2a + (n - 1) d]

S7  =   (7/2) [2 a + (7 - 1) d]

49  =  (7/2) [ 2a + 6d]

(49 x 2)/7  =  2 a + 6 d

  14  =  2 a + 6 d

  a + 3 d  =  7 -----(1)

S17  =  (17/2) [2 a + (17 - 1) d]

289 = (17/2) [ 2a + 16d]

(289 x 2)/17 = 2 a + 16 d

  34 = 2 a + 16 d

  a + 8 d = 17 -----(1)

(1) - (2)

        a + 3 d = 7

        a + 8 d = 17

       (-)  (-)    (-)

       -------------

            - 5 d = -10

                 d = 2

By applying the value of d in (1), we get 

a + 3 (2)  =  7

a + 6  =  7

a  =  7 - 6

a  =  1

To find the sum of first n terms, we have to apply the values of a and in the Sn formula

Sn  =  (n/2) [2a + (n - 1) d]

 Sn =  (n/2) [2(1) + (n - 1) (2)]

  =  (n/2) [2 + 2 n - 2]

  =  (n/2) [2 n]

  =  n²

Question 2 :

Show that a₁, a₂,............ an form an AP where an is defined as below

(i) an  = 3 + 4 n

(ii) an = 9 - 5 n

Also find the sum of 15 terms in each case.

Solution :

(i) a n = 3 + 4 n

 n  =  1

 a₁  =  3 + 4(1)

 =  7

n = 2

a₂  = 3 + 4(2)

     = 11

 d = a₂ -  a₁  =  11 - 7

d  =  4 

So the series will be in the form 7 + 11 + ...........

Now we need to find sum of 15 terms

 S15  =  (n/2) [ 2 a + (n - 1) d]

  =  (15/2) [2(7) + (15-1) 4]

  =  (15/2) [14 + 14(4)]

  =  (15/2) [14 + 56]

  =  (15/2) [70]

  =  (15 x 35)

S15  =  525

(ii) an = 9 - 5 n

Solution :

n = 1

 a1 = 9 - 5(1)  =  4

n  =  2

a2  =  9 - 5(2)  =  -1

 d  =  a2 -  a1

      -1 - 4  = -5 

So the series will be in the form 4 + (-1) + ...........

Now we need to find sum of 15 terms

 S15  =  (n/2) [ 2 a + (n - 1) d]

  = (15/2) [2(4) + (15-1) (-5)]

  =  (15/2) [8 + 14(-5)]

  =  (15/2) [8 - 70]

  =  (15/2) [-62]

  =  15 x (-31)

  =  -465

After having gone through the stuff given above, we hope that the students would have understood, sum of n terms of an arithmetic progression.

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Problems on Finding Derivative of a Function

    Mar 29, 24 12:11 AM

    Problems on Finding Derivative of a Function

    Read More

  2. How to Solve Age Problems with Ratio

    Mar 28, 24 02:01 AM

    How to Solve Age Problems with Ratio

    Read More

  3. AP Calculus BC Integration of Rational Functions by Partical Fractions

    Mar 26, 24 11:25 PM

    AP Calculus BC Integration of Rational Functions by Partical Fractions (Part - 1)

    Read More