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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.



11th grade math

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