# 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

*n*row.

^{th}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 *n ^{th}* row, then

*l*=

*a*+ (

*n*âˆ’ 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, â€¦ 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

*a*and the common difference be

*d*. Let

*S*denote the sum of first

_{n}*n*terms of the AP. Then

*S*

*=*

_{n}*a*+ (

*a*+

*d*) + (

*a*+ 2

*d*) + â€¦ + [

*a*+ (

*n*- 2)

*d*] + [(

*a*+ (

*n*-1)

*d*] (1)

The "trick" to obtain this is to rewrite

*S*as follows:

_{n}S_{n} = [*a* + (*n* - 1)* **d*] + [*a* + (*n* - 2)* **d*] + â€¦ + (*a* + *d*) + *a* (2)

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

We observe that,

Adding 1^{st} terms of both

a+ [*a* + (*n* - 1) *d*] = 2* **a* + (*n* â€“ 1)* **d*

Adding 2^{nd} terms of both

(*a* + *d*) + [*a* + (*n* â€“ 2)* **d*] = 2* **a* + (*n* â€“ 1) *d*

Adding 3^{rd} terms of both

(*a* + 2* **d*) + [*a* + (*n* â€“ 3)* d*] = 2* **a* + (*n* - 1) *d* â€¦

[*a* + (*n* â€“ 2)* **d*] + (*a* + *d*) = 2* **a* + (*n* â€“ 1) *d*

[*a* + (*n* â€“ 1)* **d*] + *a* = 2* **a* + (*n* â€“ 1)* **d*

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

2*S _{n}* =

*n*[2

*a*+ (

*n*â€“ 1)

*d*]

As a first illustration, let us verify the Gaussâ€™s answer.

Illustration

Find the sum of first 100 natural numbers.

Here *a* = 1, *d* = 1, and *n* = 100. Thus

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

(i) 16, 11, 6, â€¦ ; 23 terms, *n* terms

Here *a* = 16, *a* + *d* = 11. Therefore *d* = âˆ’ 5.

Since ,

if *n *= 23,

Also

âˆ’ 0.5, - 1.0, âˆ’ 1.5, â€¦; 10 terms, 50 terms

Here *a* = âˆ’ 0.5, *a* + *d* = âˆ’ 1.0, so that *d *= âˆ’ 0.5

Using = 5(âˆ’ 1 âˆ’ 4.5) = âˆ’ 27.5.

Also

= (25) (2 + 50 â€“ 1) (âˆ’ 0.5) = (25) (51) (-0.5) = - 637.5.

*x *+ *y*, *x* â€“ *y*, *x* â€“ 3* **y* â€¦; 22 terms, *p* terms

Here *a *= *x* + *y*, *a* + *d* = *x* â€“ *y*, so that *d* = (*x* â€“ *y*) â€“ (*x* + *y*) = âˆ’ 2* **y*.

Using , if *n* = 22 we get

= (11) [2

*x*+ 2

*y*â€“ 42

*y*]

= (11) [2

*x*â€“ 40

*y*]

= 22 (

*x*â€“ 20

*y*)

and

= *p* [*x* + (2 â€“ *p*)* **y*].

1, 3, 5, 7, â€¦; 100 terms, 200 terms

Here *a* = 1, *a* + *d* = 3, so that *d* = 2.

Using (When *n* = 100)

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

and (When *n* = 200)

Here

Using , we get

and

.

Here .

Using , if n = 10 we get

and .

Here

Using

0.9, 0.91, 0.92, 0.93, â€¦; 20 terms, *n *terms

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 .

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

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 + â€¦ + 399

Here *t*_{1} = *a* = 203, *d* = 7, *l* = 399

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

Then 399 = 203 + (*n *- 1) (7)

â‡’7* n* = 399 â€“ 203 + 7 = 406 â€“ 203 = 203

*n*= 29.

Hence the required sum

= (29) (301) = 8729.

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

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 + â€¦ + 999.

Here *t*_{1} = *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)

â‡’ 111= 12 + (*n* - 1) (Dividing by 9 on both sides)

â‡’ *n *= 100

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

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

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 + â€¦ + 997

Here *t*_{1} = *a* = 101, *d* = 4 and *l* = 997.

Let *n* denote the number of terms in the series.

*n*â€“ 1)

â‡’ 896 = 4 (*n* â€“ 1)

â‡’ 900 = 4* n*

*n* = 225.

Hence, the required sum

= (225) (549) = 123525.

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

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, â€¦, 245 (1)

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

245 = 7 + (*n *â€“ 1) (7)

â‡’ 7* n* = 245

â‡’ *n* = 35

Sum of the numbers in (1)

.

The numbers which are divisible by 4 are

4, 8, 12, â€¦, 248 (2)

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

248 = 4 + (*m* â€“ 1) (4)

â‡’ 248 = 4

*m*â‡’

*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, â€¦, 224 (3)

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

224 = 28 + (*r* â€“ 1) (28)

â‡’ *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 â€“ 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 â€“ 11214 = 20161.

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

The positive integers which are divisible by 5 are 5, 10, 15, 20, â€¦, 1000

Out of these 10, 20, 30, â€¦ 1000 are divisible by 2.

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

5, 15, 25, â€¦, 995 (1)

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

995 = 5 + 10 (*n *â€“ 1)

â‡’ 1000 = 10* **n*

*n *= 100.

Thus, the sum of numbers in (1)

.

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

Let the two AP's be a, *a* + *d*, *a *+ 2* **d*, *a* + 3* **d*, â€¦ (1)

and *A*, *A* + *D*, *A* + 2* **D*, *A* + 3* **D*, â€¦ (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 7^{th }term

We therefore put or

*n*= 13 in (3), to obtain

.

â‡’ *a* + 6* **d*, : *A* + 6*D* = 40 : 57.

If the sum of *p* terms an AP to the sum of *q* terms is *p*^{2 }: *q*^{2}. Show that the common difference is twice the first term and that the ratio of the *p ^{th}* term to the

*q*

*term is 2*

^{th}

*p*â€“ 1 : 2

*q*â€“ 1. (Assume

*q*â‰

*p*.)

Let the AP be a

,*a*+

*d*,

*a*+ 2

*d*,

*a*+ 3

*d*, â€¦

We are given that

â‡’ 2* **a* *q* + *p* *q d* â€“ *q d* = 2* **a* *p* + *p* *q d* - *p d*

â‡’2* **a* (*q* â€“ *p*) = (*q* â€“ *p*) *d* [cancel *pdq* from both the sides]

â‡’ 2

*a*=

*d*[Q

*q*â‰

*p*]

Also, (By substituting 2

*a*=

*d*)

â‡’*t _{p}* :

*t*

_{q}= 2

*p*â€“ 1 : 2

*q*â€“ 1.

If *S*_{1}, *S* _{2}, *S* _{3}, â€¦ *S*_{q} are the sums of n terms of *q*. Arithmetic Progressions whose first terms are 1, 2, 3, â€¦, *q* and common differences are 1, 3, 5, â€¦, (2*q* - 1) respectively, show that

The *r*^{th} series is an AP with the first term as *r *and common difference as (2*r* â€“ 1).