Question

If A be the first term and D
be common difference of an A.P.
Prove that S =" [2a +(n-1)d]

Answer 1

Answer:

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]

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Answer 2

Answer:

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]