Question

prove that the sum of first terms of an A.P is
s=n/2[2a+(n-1)d]
where first term is a
and the common difference is d

plz if u dont know the answer plz leave it and who can explain me properly he will be awardedas a a brainlist...

Answer 1

Suppose, S denote the sum of the first n terms of the Arithmetic Progression {a, a + d, a + 2d, a + 3d, a + 4d, a + 5d ……………...}.

Now nth term of the given Arithmetic Progression is a + (n - 1)d

Let the nth term of the given Arithmetic Progression = l

Therefore, a + (n - 1)d = l

Hence, the term preceding the last term is l – d.

The term preceding the term (l - d) is l - 2d and so on.

Therefore, S = a + (a + d) + (a + 2d) + (a + 3d) + …………………….. to n tems

Or, S = a + (a + d) + (a + 2d) + (a + 3d) + …………………….. + (l - 2d) + (l - d) + l ……………… (i)

Writing the above series in reverse order, we get

S = l + (l - d) + (l - 2d) + ……………. + (a + 2d) + (a + d) + a………………(ii) 

Adding the corresponding terms of (i) and (ii), we get

2S = (a + l) + (a + l) + (a + l) + ……………………. to n terms

⇒ 2S = n(a + l)

⇒ S = n/2(a + l)

⇒ S = Numberofterms/2 (First term + Last term) …………(iii)

⇒ S = n/2[a + a + (n - 1)d], Since last term l = a + (n - 1)d

⇒ S = n/2[2a + (n - 1)d]

Answer 2

The general series is S=a+(a+d)+(a+2d)+(a+3d) +....+(l-d))+l where a is the first term,d is the common difference and l is the last term. 
If we write this in reverse we get S=l+(l-d)+(l-2d)+(l-3d)+....+(a+d)+a. 
So we have,
S=a+(a+d)+(a+2d)+(a+3d)+....+(l-d)+l (i)
and
S=l+(l-d)+(l-2d)+(l-3d)+....+(a+d)+a (ii) 
Now add (i) and (ii) together, making sure that you add corresponding terms together and you get
2S=(a+l)+(a+l)+(a+l)....+(a+l)+(a+l).
And so
2S= n(a+l) (because there are n lots of (a+l))
So
S= [n/2(a+l)] (iii) 
But your last term or nth term can be written as a+(n-1)d so
l=a+(n-1)d
Now replace l in (iii) and we get
S=[n/2(a+a+(n-1)d)]
This simplifies to 
S=[n/2(2a+(n-1)d)]