untitled

or, 2Sn n ap [number of terms of the series is n]

? ( )

2

a p iii

n

Sn 

Again,nth term p a(n 1 )d. Putting this value of p in (iii) we get,

[ { ( 1 ) }]

2

a a n d

n

Sn   

i.e., ^` 2  1 ( )
2

a n d iv

n

Sn  

If the first term of arithmetic series a, last term p and number of terms n are
known, the sum of the series can be determined by the formula (iii). But if first term
thea, common difference d,number of terms n are known, the sum of the series are
determined by the formula (iv).
Determination of the sum of first n terms of natural numbers
Let Snbe the sum of n-numbers of natural numbers i.e.
Sn 1  2  3 (n 1 )n (i)
Writing from the first term and conversely fr om the last term of the series we get,
Sn 1  2  3  n 2 (n 1 )n (i)
and Sn n n 1 n 2  3  2  1 (ii)
Adding, 2 Sn (n 1 )(n 1 )(n 1 )(n 1 )[n-number of terms]
or, 2 Sn n(n 1 )

? ( )

2

( 1 )

iii

nn

Sn



Example 2. Find the sum total of first 5 0 terms of natural numbers.
Solution : Using formula ( iii) we get,

25 51 1275

2

50 ( 50 1 )

50 u^



S

? The sum total of first 50 natural numbers is 1275.
Example 3. 1  2  3  4  99 what?
Solution : The first term of the series a 1 , common difference d 2  1 1 and
the last termp = 99.
? It is an arithmetic series.
Let the nth term of the series = 99
We know, nth term of an arithmetic
progression =a(n 1 )d
? a(n 1 )d 99
or, 1 (n 1 ) 1 99
or, 1 n 1 99

Alternative method :

Since

,

2

a p

n

Sn 

? 1 99

2

99

S 99