# Sum of arithmetic progression

The sum of arithmetic progression is denoted by $S_{n}$. It is nothing but the sum of 'n terms of an A.P. with first term 'a' and common difference 'd'.The formula for sum of n terms of A.P. is

$S_{n} = \frac{n}{2}[2a + (n-1)d]$

$S_{n} = \frac{n}{2}[a + l]$ , where l = last term = a + (n -1 )d

**Proof :**Let $a_{1}, a_{2},a_{3},...,a_{n}$ be an A.P. with first term as 'a' and common difference as 'd'.

$a_{1}$ = a; $a_{2}$ = a + d; $a_{3}$ = a + 2d ; ... $a_{n}$ = a + (n -1)d

$S_{n} = a_{1} + a_{2} + a_{3} + ... +a_{n-1} + a_{n}$

⇒ $S_{n}$ = a + (a + d) + (a + 2d) + … + [ a + (n-2)d] + [ a + (n -1 )d] -----(i)

Write the above equation in reverse order we get,

$S_{n}$ = [ a + (n -1 )d] + [ a + (n-2)d] + … + (a + 2d) + (a + d) + a ----- (ii)

Now add the two equations,

2$S_{n}$ = [ 2a + (n -1 )d] + [ 2a + (n-1)d] + …+ [2a + (n-1)d]

[2a + (n-1)d] repeats ‘n’ times

∴ 2$S_{n}$ = n [ 2a + (n -1 )d]

$S_{n} = \frac{n}{2} $ [ 2a + (n -1 )d]

Since the last term l = a + (n – 1)d

∴ $S_{n} = \frac{n}{2} $ [ a + a + (n -1 )d]

$S_{n} = \frac{n}{2} $ [a + l ]

**Note :**In the above formula there are 4 unknown quantities. So if any three are given then we can find the forth one.

In the sum of $S_{n}$ of n terms of a sequence is given then the nth term $a_{n}$ of the sequence can determined by using the following formula .

$a_{n} = S_{n} - S_{n - 1}$

## Solved examples on sum of arithmetic progression

1) 50,46,42,… 10 terms.**Solution:**50,46,42,… 10 terms

The formula to find sum is

$S_{n} = \frac{n}{2} $ [ 2a + (n -1 )d]

Number of terms = n = 10; First term = a = 50 ; Common difference = d = 46 – 50 = -4

Put all the given values in the formula we get,

$S_{10} = \frac{10}{2} [ 2 \times$ 50 + (10 -1 )(-4)]

$S_{10}$ = 5 [ 100 + (9 )(-4)]

$S_{10}$ = 5 [ 100 + (-36)]

$S_{10}$ = 5 (64)

$S_{10}$ = 320

2) 3, $\frac{9}{2}$, 6, $\frac{15}{2} $ ,… 25 terms .

**Solution:**3, $\frac{9}{2}$, 6, $\frac{15}{2} $ ,… 25 terms .

The formula to find sum is

$S_{n} = \frac{n}{2} $ [ 2a + (n -1 )d]

Number of terms = n = 25; First term = a = 3 ; Common difference = d = , $\frac{9}{2}$ - 3 = $\frac{3}{2}$

Put all the given values in the formula we get,

$S_{25} = \frac{25}{2} [ 2 \times 3 + (25 -1) \frac{3}{2}$]

$S_{25}$ = 12.5 [ 6 + (24 )(1.5)]

$S_{25}$ = 12.5 [ 6 + 36]

$S_{25}$ = 12.5 (42)

$S_{25}$ = 525

3) In an A.P. the sum of first n terms is $ \frac{3n^{2}}{2} + \frac{13}{2}n$. Find its 25th term.

**Solution :**$S_{n} = \frac{3n^{2}}{2} + \frac{13}{2}n$.

When n = 25,

$S_{25} = \frac{3 \times 25^{2}}{2} + \frac{13}{2} \times25$.

$S_{25}$ = 1100

Now we will find $S_{n - 1} = S_{25 -1} =S_{24}$

$S_{24} = \frac{3 \times 24^{2}}{2} + \frac{13}{2} \times24$.

$S_{24}$ = 1080

As we know that ,

$a_{n} = S_{n} - S_{n -1}$

$a_{25} = S_{25} - S_{24}$

$a_{25}$ = 1100 – 1080 = 80

∴ 25th term is 80.

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