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Sum of First n Terms of an AP

Let us return to our auditorium of first section and try to answer the following question.

How many seats are there in an auditorium in which the front row has

a seats; each row (other than the front row) has d more seats than the row in front of it?

Schematically, the auditorium looks like the figure given below. For convenience let

l denote the number of seats in the last or nth row.

There are n rows. The front row has a seats. The last row has l seats. Let S be the total number of seats.

 


Now copy the diagram, turn it upside down, and place it next to the given diagram.

 

 

Two copies of the auditorium produces a figure consisting of

n rows each having a + l seats.

If S is the sum of the seats in the auditorium, then 2 S = n (a + l)

If the last row is the nth row, then l = a + (n − 1) d and the above formula becomes



This gives us the sum to n terms of an AP.


To derive this formula algebraically, we begin with the following well authenticated story about Karl Friedrich Gauss.


In his first arithmetic class, Gauss astonished his teacher by instantly solving what was intended to be a "busy work" problem:


Find the sum of all the natural numbers from 1 to 100.

Gauss recognized the pattern 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, … 50 + 51 = 101.

Since there are 50 pairs of numbers, each of which adds up to 101, the sum of all the numbers must be (50) (101) = 5050.


Let us use a technique similar to the one used by Gauss to find the sum of n terms of an AP.

Sum to n terms of an AP

Let the first term of an AP be a and the common difference be d. Let S n denote the sum of first n terms of the AP. Then Sn = a + (a + d) + (a + 2 d) + … + [a + (n - 2) d] + [(a + (n -1) d]           (1)


The "trick" to obtain this is to rewrite

Sn as follows:

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


we add the terms of (1) and (2) vertically.

We observe that,


Adding 1st terms of both

a+ [a + (n - 1) d] = 2 a + (n – 1) d


Adding 2nd terms of both

(a + d) + [a + (n – 2) d] = 2 a + (n – 1) d


Adding 3rd terms of both

(a + 2 d) + [a + (n – 3) d] = 2 a + (n - 1) d   …

[a + (n – 2) d] + (a + d) = 2 a + (n – 1) d

[a + (n – 1) d] + a = 2 a + (n – 1) d


Since there are n terms in each of the equations (1) and (2), adding both the equations we get,

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


As a first illustration, let us verify the Gauss’s answer.

Illustration
 

Example

Find the sum of first 100 natural numbers.

Solution

Here a = 1, d = 1, and n = 100. Thus


 

Example

 Find the sum of indicated number of terms of each of the following arithmeticprogressions:

 (i) 16, 11, 6, … ; 23 terms, n terms

Solution

Here a = 16, a + d = 11. Therefore d = − 5.

Since

if n = 23,

Also

 

 

Example

− 0.5, - 1.0, − 1.5, …; 10 terms, 50 terms

Solution

Here a = − 0.5, a + d = − 1.0, so that d = − 0.5

Using = 5(− 1 − 4.5) = − 27.5.

Also

            = (25) (2 + 50 – 1) (− 0.5) = (25) (51) (-0.5) = - 637.5.


 

Example

x yx – yx – 3 y …; 22 terms, p terms

Solution

 Here a x + ya + d = x – y, so that d = (x – y) – (x + y) = − 2 y.

Using , if n = 22 we get

 

      = (11) [2

x + 2 y – 42 y]

      = (11) [2

x – 40 y]

      = 22 (

x – 20 y)

and  

           = p [x + (2 – p) y].


 

Example

1, 3, 5, 7, …; 100 terms, 200 terms

Solution

Here a = 1, a + d = 3, so that d = 2.

Using                 (When n = 100)

             = (50) (2 + 198) = (50) (200) = 10000

and                         (When n = 200)

 

 

Example

 

Solution

Here

Using , we get

                         

and

          .


 

Example

 

Solution

Here .

Using ,  if n = 10 we get

and .


 

Example

 

Solution

 Here  

Using  

 

 

Example

0.9, 0.91, 0.92, 0.93, …; 20 terms, n terms

Solution

 Here a = 0.9, a + d = 0.91, so that d = 0.01

Using

             = 10[1.8 + (19) (0.01)] = 10 [1.8 + 0.19] = 10 (1.99) = 19.9

and

 

 

Example

Find the sum of all integers which are multiples of 7 and lie between 200 and 400.

Solution

Since 4 is left as remainder when we divide 200 by 7 the least number greater than 200 divisible by 7 is 203. Again if we divide 400 by 7, 1 is the left as remainder. This implies that the greatest number less than 400, which is divisible by 7 is 399. Thus, we have to find the sum in the series

203 + 210 + 217 + … + 399

Here t1 = a = 203, d = 7, l = 399

Let n be the total number of terms in this series. 
Then 399 = 203 + (
n - 1) (7)

7 n = 399 – 203 + 7 = 406 – 203 = 203  

 ⇒   n = 29.

Hence the required sum

                                 = (29) (301) = 8729.


 

Example

Find the sum of all the integers between 100 and 1000 which are divisible by 9.

Solution

The first integer greater than 100 and divisible by 9 is 108 and the integer just smaller than 1000 and divisible by 9 is 999. Thus, we have to find the sum of the series

108 + 117 + 126 + … + 999.

Here t1 = a = 108, d = 9 and l = 999

Let n be the total number of terms in the series be n. Then

999 =108 + 9 (n - 1)   

⇒ 111= 12 + (n - 1)    (Dividing by 9 on both sides)

   n = 100

Hence, the required sum = 50 (1107) = 55350.


 

Example

Find the sum of all 3 digit numbers which leaves remainder 1 when divided by 4.

Solution

 The smallest 3-digit number is 100. It is divisible by 4, therefore, the first 3-digit number which when divided by 4 leaves remainder 1 is 101. The largest 3-digit number is 999 which, when divided by 4 leaves remainder 3. Thus the largest 3-digit number which when divided by 4 leaves remainder 1 is 997. Thus, we have to find the sum of

101 + 105 + … + 997

Here t1 = a = 101, d = 4 and l = 997.

Let n denote the number of terms in the series. 

Then 997 = 101 + 4 (n – 1)

     896 = 4 (n – 1)

    ⇒ 900 = 4 n

            n = 225.

Hence, the required sum

                                  = (225) (549) = 123525.

 

 

Example

Find the sum of all integers from 1 to 250, excluding, those which are divisible either by 7 or by 4.

Solution

 Let us find the sum of all the integers between 1 and 250 which are divisible either by 7 or by 4.

The numbers which are divisible by 7 are

7, 14, 21, …, 245                                     (1)

If n is the number of terms in this sequence, then

    245 = 7 + (n – 1) (7)

   7 n = 245

    ⇒ n = 35

Sum of the numbers in (1)

            .

The numbers which are divisible by 4 are

4, 8, 12, …, 248                                       (2)

If m is the number of terms in (2), then

       248 = 4 + (m – 1) (4)

   

 248 = 4 m  m = 62.

Sum of the numbers in (2) .

The sequence in (1) and (2) have the following common terms (the terms which are divisible both by 7 and 4, i.e. are divisible by 28)

28, 56, …, 224                                       (3)

If r is the number of terms in (3), then

        224 = 28 + (r – 1) (28)

        r = 224/28 = 8.

Sum of the numbers in (3) .

The sum of the numbers which are divisible either by 7 or by 4 (see in the figure given below).

        = 4410 + 7812 – 1008 = 11214.(as the numbers given by (3) will be repeated in (1) and (2))

Also, sum of the integers from 1 to 250                                   

        .

Hence, sum of the integers from 1 to 250 excluding those which are divisible either by 7 or by 4 = 31375 – 11214 = 20161.

 


 

Example

Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2.

Solution

The positive integers which are divisible by 5 are 5, 10, 15, 20, …, 1000

Out of these 10, 20, 30, … 1000 are divisible by 2.

Thus, we have to find the sum of the positive integers

5, 15, 25, …, 995                                               (1)

If n is the number of terms in it the sequence (1), then

 

            995 = 5 + 10 (n – 1)

      ⇒ 1000 = 10 n

               n = 100.

Thus, the sum of numbers in (1)

          .


 

Example

The ratio of the sums of n terms of two AP is 3 n + 1 : 5 n – 8; find the ratio of their 7th terms.

Solution

Let the two AP's be a, a + da + 2 da + 3 d, …                                            (1)

and AA + DA + 2 DA + 3 D, …                                     (2)

Sum to n terms of (1) is  and sum to n terms of (2) is .

We are given that

                 (3)

We are looking at the 7th term

We therefore put or

n = 13 in (3), to obtain

.

⇒ a + 6 d, : A + 6D = 40 : 57.

 

 

Example

If the sum of p terms an AP to the sum of q terms is p2 q2. Show that the common difference is twice the first term and that the ratio of the pth term to the qth term is 2 p – 1 : 2 q – 1. (Assume q ≠ p.)

Solution

Let the AP be a

a + da + 2 da + 3 d, …

We are given that

⇒ 2 a q + p q d – q d = 2 a p + p q d - p d

2 a (q – p) = (q – pd      [cancel pdq from both the sides]

           ⇒ 2

a = d                         [Q q ≠ p]

Also,        (By substituting 2

a = d)

tp : tq = 2 p – 1 : 2 q – 1.


 

Example

If S1S 2S 3, … Sq are the sums of n terms of q. Arithmetic Progressions whose first terms are 1, 2, 3, …, q and common differences are 1, 3, 5, …, (2q - 1) respectively, show that

Solution

The rth series is an AP with the first term as r and common difference as (2r – 1).

  

 





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