Math, asked by havvagb786, 9 months ago

if algebraic form of an arthimetic squence is xn=an+b then sum of first n terms sn=an(n+1)by2+bn​

Answers

Answered by kaustumbh3136
1

Answer:Prove that the sum Sn of n terms of an Arithmetic Progress (A.P.) whose first term ‘a’ and common difference ‘d’ is

S = n2[2a + (n - 1)d]

Or, S = n2[a + l], where l = last term = a + (n - 1)d

Proof:

Suppose, a1, a2, a3, ……….. be an  Arithmetic Progression whose first term is a and common difference is d.

Then,

a1 = a

a2 = a + d

a3 = a + 2d

a4 = a + 3d

………..

………..

an = a + (n - 1)d

Now,

S = a1 + a2 + a3 + ………….. + an−1 + an

S = a + (a + d) + (a + 2d) + (a + 3d) + ……….. + {a + (n - 2)d} + {a + (n - 1)d} ……………….. (i)

By writing the terms of S in the reverse order, we get,S = {a + (n - 1)d} + {a + (n - 2)d} + {a + (n - 3)d} + ……….. + (a + 3d) + (a + 2d) + (a + d) + a

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

2S = {2a + (n - 1)d} + {2a + (n - 1)d} + {2a + (n - 1)d} + ………. + {a + (n - 2)d}

2S = n[2a + (n -1)d

⇒ S = n2[2a + (n - 1)d]

Now, l = last term = nth term = a + (n - 1)d

Therefore, S = n2[2a + (n - 1)d] = n2[a {a + (n - 1)d}] = n2[a + l].

 

We can also find find the sum of first n terms of an Arithmetic Progression according to the process below.

Step-by-step explanation:

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