Question

What is the formula for sum of n terms of a Arithmetic Progression? Explain each term.

Answer 1

standard form of arithmetic progression is given as ....

a , (a + d) , (a + 2d), (a + 3d), (a + 4d) , ..... $\bf{a_n}$

we have to find sum of n terms of given arithmetic progression.

i.e., $S_n$ = a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + ....... + $a_n$

as you know, nth term of an AP , $a_n$ = a + (n - 1)d

so, $S_n$ = a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + ....... + {a + (n - 1)d}

= (a + a + a + ..... n times) + {d + 2d + 3d + 4d + .... (n - 1)d }

= na + d {1 + 2 + 3 + 4 + .... + (n - 1)}

= na + d{ n(n - 1)/2 }

[ we know, sum of n natural number is n(n + 1)/2, so. sum of (n - 1) natural number is n(n - 1)/2 . ]

$S_n$ = n [a + d(n -1)/2 ]

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

hence, sum of n terms of an ap is given as Sn = n/2 [2a + (n - 1)d ]

Answer 2

DEFINITION

ARITHMETIC PROGRESSION

An Arithmetic Progression is a sequence of numbers in which we get each term of the sequence ( except first term ) is obtained by adding a particular number to the previous term.

NOTATIONS

The constant term which is to added is called COMMON DIFFERENCE. It is denoted by d

The FIRST TERM is denoted by a

TO DETERMINE

The formula for sum of n terms of a Arithmetic Progression

CALCULATION

The progression upto n terms is

So the sum is

$\sf{a + (a+d) + (a+2d) + ......... + \{a+(n-1)d \} \: }$

$= \sf{na + \{ 1 + 2 + 3 + .... + (n - 1)\ \}d \: \: }$

$\displaystyle \: = \sf{na +\: \frac{(n - 1)(n - 1 + 1)}{2} d \: }$

$\displaystyle \: = \sf{na +\: \frac{n(n - 1)}{2} d \: }$

$\displaystyle \: = \sf{ \frac{n}{2} \{2a +\: (n - 1)d \} }$

Which is the required formula

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