Sum of First n Terms of an Arithmetic Progression Class 10th

Introduction

We know that there are n terms in an Arithmetic Progression (AP) and if we want the sum of the first n terms of an Arithmetic Progression then what should we do, we shall add all the terms of the AP but that will take more time to solve. Sometimes this method will not give the answer, to add first n terms of an Arithmetic Progression there is a formula for an easy solution.

Before the formula, let’s take an example to understand it clearly.

Explanation

I gave some money to my friend to help him and he gave me the money in instalments like in the first month he gave me 1000 rupees, in the second month he gave me 2000 rupees and so on. He gave all the money in 15 months in this order. Now I want to calculate that after 15 months how much money did I get? So I shall have to add all the instalments but that will take too much time. How can I add all the instalments in a shorter way? There is a formula to add easily. How can we get the formula, let’s see?

If I write all the instalments to get a sum and the sum is denoted by S

S = 1000 + 2000 + 3000 +…………………+ 13000 + 14000 + 15000

Now let’s rewrite all the terms in reverse order

S = 15000 + 14000 + 13000 +…………….…+ 3000 + 2000 + 1000

Now adding both the series

S + S = (1000 + 15000) + (2000 + 14000) + (3000 + 13000) + …….+ (13000 + 3000) + (14000 + 2000) + (15000 + 1000)

2S = 16000 + 16000 + 16000 +……….+ 16000 + 16000 + 16000 (15 times)

Therefore, 2S = 16000⨯15

S = 16000⨯15/2 = 8000⨯15

S = 120000

it means I shall get 1,20,000 rupees after 15 months.

Similarly, same as the above method, we can generate the formula to find the sum of the first n terms of an Arithmetic Progression.

Derivation of the Formula

We write the AP with first term a and common difference d for n terms as follows

a, a + d, a + 2d +………..+ a + (n – 1)d 

The sum of the first n terms of the Arithmetic Progression is denoted by Sn, so we can write

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

Let’s rewrite all the terms in reverse order

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

Now, adding both the equations (1) and (2)

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

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

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

2Sn = [2a + (n – 1)d] + [2a + (n – 1)d] +……………..+ [2a + (n – 1)d] + [2a + (n – 1)d] {for n times}

2Sn =  [2a + (n – 1)d]⨯n

Sn = [2a + (n – 1)d]⨯n/2

Sn = n/2[2a + (n – 1)d]

So, the sum of the first n terms of an Arithmetic Progression        Sn = n/2[2a + (n – 1)d] 

Where Sn = Sum of First n terms of an AP

n = Number of Terms

a = First Term

d = Common Difference

SUM OF FIRST n TERMS OF AN ARITHMETIC PROGRESSION

This can also be written as Sn = n/2[a + a + (n – 1)d]

Sn = n/2[a + an]                [∵ nth term an = a + (n – 1)d]

If there are only n terms in AP then an = l (last term)

Therefore,       Sn = n/2[a + l]

If the first term (a) and the last term ( l ) are given then the above formula is useful.

If we want to solve the above-solved example with the help of a formula then first term (a) = 1000, last term ( l ) = 15000, number of terms (n) = 15

So, the sum will be S15 = 15/2[1000 + 15000]

S15 = 15/2[16000]

S15 = 15⨯8000

S15 = 120000

We can see that we have solved it very conveniently with the help of a formula.

Note – The nth term of an AP is equal to the difference of the sum of first n terms and the sum of first (n – 1) terms.

i.e.        an = Sn – Sn – 1        

Let’s prove it.

RHS = Sn – Sn – 1

= n/2[2a + (n – 1)d] – (n-1)/2[2a + {(n – 1) – 1}d]

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

= 1/2[2an + n2d – nd] – 1/2[2a(n – 1) + nd(n – 1) – 2d(n – 1)]

= 1/2[2an + n2d – nd – 2a(n – 1) – nd(n – 1) + 2d(n – 1)]

= 1/2[2an + n2d – nd – 2an + 2a – n2d + nd + 2dn – 2d]

= 1/2[2a + 2dn – 2d]

= 2/2[a + dn – d]

= a + (n – 1)d

= an = LHS                    Hence Proved.

Some Examples –

Example 1) Find the sum of first 16 terms of the AP: 1, 10, 19, 28………….

Solution – Here first term (a) = 1, common difference (d) = 10 – 1 = 9

Number of terms (n) = 16,   S16 = ?

By the formula of sum of first n terms, Sn = n/2[2a + (n – 1)d]

Putting the values       S16 = 16/2[2⨯1 + (16 – 1)9]

S16 = 8[2 + 15⨯9]

S16 = 8[2 + 135]

S16 = 8[137]

S16 = 1096

So, the sum of the first 16 terms is 1096.               Ans.

Example 2) Find the sum of the series: 10 + 7 + 4 + 1 +………..+ (-80).

Solution – Here common difference (d) = 7 – 10 = -3

4 – 7 = -3

1 – 4 = -3

The common difference is the same so it is an AP.

First term (a) = 10, last term an = l = -80, number of terms (n) = ?

Sum of n terms (Sn) =?

So, first we shall find the number of terms (n)

By the nth term formula,    an = a + (n – 1)d

-80 = 10 + (n – 1)(-3)

-80 – 10 = -3n + 3

-90 = -3n + 3

3n = 90 + 3

n = 93/3 ⇒ n = 31

It means there are 31 terms in the given series.

So, sum of first 31 terms of the AP by the formula,   Sn = n/2[a + l ]

S31 = 31/2[10 + (-80)]

S31 = 31/2[10 – 80]

S31 = 31/2[-70] = 31⨯(-35)

S31 = -1085

Therefore, the sum of the first 31 terms is -1085.              Ans.

Example 3) How many terms of the AP: 19, 21, 23, 25,……….must be taken so that their sum is 115.

Solution – First term (a) = 19, common difference (d) = 21 – 19 = 2,   Sn = 115

The number of terms (n) =?

Since,   Sn = n/2[2a + (n – 1)d]

115 = n/2[2⨯19 + (n – 1)2]

115⨯2 = n[38 + 2n – 2]

230 = 38n + 2n2 – 2n

2n2 + 36n – 230 = 0

2[n2 + 18n – 115] = 0

n2 + 18n – 115 = 0

By factorization method,       n2 + 23n – 5n – 115 = 0

n(n + 23) – 5(n + 23) = 0

(n + 23)(n – 5) = 0

Now, n + 23 = 0         and         n – 5 = 0

n = -23        and         n = 5

Since we know that number of terms (n) can’t be negative.

So, n = 5 is the required number of terms.

It means 5 terms must be taken so that their sum is 115.            Ans.

Example 4) Find the sum of the first n positive integers.

Solution – We know that the positive integers are   1, 2, 3, 4,………………..n

Here first term (a) = 1, last term ( l ) = n,   Sn = ?

Here the last term is given so we shall use the formula Sn = n/2[a + l ]

Putting the values, Sn = n/2[1 + n]

Or we can write,    Sn = n(n+1)/2 

This can be used as the formula to find the sum of positive integers if the number of terms (n) is given.

Example 5) Find the sum of the first 10 terms of AP whose nth term is an = 7n – 9.

Solution – Here nth term    an = 7n – 9

First we shall generate the AP by nth term

For n = 1,         a1 = 7⨯1 – 9 = 7 – 9 = -2

For n = 2,         a2 = 7⨯2 – 9 = 14 – 9 = 5

For n = 3,         a3 = 7⨯3 – 9 = 21 – 9 = 12

So, the AP is -2, 5, 12,…………………

first term (a) = -2, common difference (d) = 5 – (-2) = 5 + 2 = 7

Now by the formula,    Sn = n/2[2a + (n – 1)d]

Sum of first 10 terms,  S10 = 10/2[2⨯(-2) + (10 – 1)7]

S10 = 5[-4 + (9)7]

S10 = 5[-4 + 63]

S10 = 5[59]

S10 = 295

Therefore, the sum of the first 10 terms of the given AP is 295.             Ans.

Example 6) Find the sum of natural numbers that fall between 350 to 1000 which is divisible by 4.

Solution – Between 350 to 1000, numbers divisible by 4 are 352, 356, 360,…………….1000.

This series is an AP with first term (a) = 352, common difference (d) = 356 – 352 = 4

Last term (an) = ( l ) = 1000, number of terms (n) = ?

To find sum we need number of terms (n) so first we shall find it by nth term formula,

an = a + (n – 1)d

1000 = 352 + (n – 1)4

1000 – 352 = 4n – 4

648 + 4 = 4n

652 = 4n

652/4 = n

n = 163

It means there are 163 terms between 350 to 1000 which are divisible by 4.

Now, sum of 163 terms by the formula,    Sn = n/2[a + l ]

S163 = 163/2[352 + 1000]

S163 = 163/2[1352] = 163⨯676

S163 = 110188

So, the sum of 163 terms which is divisible by 4 is 110188.              Ans.

Example 7) Find the first term of an AP whose sum of n terms is given by 3n2 – 2n. Also find the sum of the first two terms, 2nd term, 3rd term, 11th term and nth term.

Solution – Here sum of n terms of an AP,     Sn = 3n2 – 2n

For n = 1,     S1 = 3(1)2 – 2⨯1 = 3 – 2 = 1

So, first term (a) = 1

Now sum of first two terms putting n = 2,  S2 = 3(2)2 – 2⨯2 = 3(4) – 4 = 12 – 4 = 8

So, the sum of the first two terms is 8.

Second term a2 = S2 – S1            [by the formula, an = Sn – Sn – 1]

a2 = 8 – 1

a2 = 7

So, the 2nd term is 7.

In the question, it is given AP so the common difference will be the same.

Common difference of first two terms d = a2 – a1 = 7 – 1 = 6

So, the 3rd term will be       a3 = a2 + d = 7 + 6 = 13

3rd term       a3 = 13

Now the AP: 1, 7, 13,…………..

So, nth term    an = a + (n – 1)d

an = 1 + (n – 1)6

an = 1 + 6n – 6

nth term    an = 6n – 5 

For n = 11,    a11 = 6⨯11 – 5

a11 = 66 – 5

So,  11th term      a11 = 61              Ans.

Sum of First n Terms of an Arithmetic Progression Class 10th in Hindi

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