Question

Derive the formula for the nth term of an A.P and sum of first n term of an A.p

Answer 1

Step-by-step explanation:

We know a, a+d, a + 2d ,... is an A.P

where a -> first term & d is the common difference given by (Tn - Tn-1)

if T1 is the first term , then we can assume that Tn is the n'th term.

so,

T1 = a        = a + (1-1) d

T2 = a + d = a + (2-1) d

T3 = a + 2d = a + (3-1) d     and so on......

similiarly,

Tn = ..............= a + (n-1)d  , which is the formula for n'th term of an A.P

and next to summation formula,

let us assume Sn be the sum of A.P up to n'terms

then,

Sn = a + (a+d) + (a+2d) + (a + 3d) + ...... + [a +(n-1)d] ------------- 1

if we assume 'l' as the last term, then we can rewrite the same Sn as,

Sn = l + (l-d) + (l-2d) + (l-3d) + ........... + [l + (n-1)d] -------------------2

add 1 and 2,

2*Sn = [ n*a + n*l ]

Sn = (n/2) [ a + l ]

which can be again rewritten as,

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

ie.,

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