Question

From what value of "a"of sequence 3a+24,4a+2aand 6a+31 from ap

Answer 1

Answer:

,

Step-by-step explanation:

find the sum of first n terms of an Arithmetic Progression.

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.

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]