soon w The sum of first in terms of an A.P, whose first term is 'aand last term is 1 A) Sa+(n+1)d) B) S. -- 11 C) 5122*(n+1)] DS
Answers
Answer:
ove 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.
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 = n2(a + l)
⇒ S = Numberofterms2 × (First term + Last term) …………(iii)
⇒ S = n2[a + a + (n - 1)d], Since last term l = a + (n - 1)d
⇒ S = n2[2a + (n - 1)d]
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