Question

derive the formula of the sum of first n terms of an ap whose first term is a and common difference is d​

Answer 1

Then,

a11 = a

a22 = a + d

a33 = a + 2d

a44 = a + 3d

………..

………..

ann = a + (n - 1)d

Now,

S = a11 + a22 + a33 + ………….. + an−1n−1 + ann

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 = n2n2[2a + (n - 1)d]

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

Therefore, S = n2n2[2a + (n - 1)d] = n2n2[a {a + (n - 1)d}] =n2n2[a + l].

 

Answer 2

Step-by-step explanation:

HELLO MATE

IN AN AP OF N TERMS

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

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