• If three terms are in A.P., then the middle term is called the arithmetic mean (A.M.) between the other two i.e. if a, b, c are in A.P. then b = a+c/2 is the A.M. of a and c.
• If a1, a2, ……, an are n numbers, then the arithmetic mean (A) of these numbers is A = 1/n (a1 + a2 + a3 +…+ an).
• The n numbers A1, A2, ……, An are said to be A.M.’s between the numbers a and b if a, A1, A2, ……, An, b are in A.P. If d is the common difference of this A.P., then b = a + (n + 2 – 1)d ⇒ d = b–a/n+1.
⇒ A1 = a + b–a/n+1 = na+b A2 = a + 2(b–a)/n+1 ,…, An = a + n(b–a)/n+1 = a+nb/n+1.
Illustration:
If 1st and 2nd terms of an A.P. are 1 and –3 respectively, find the nth term and the sum of the first n terms.
Solution:
1st term = a, 2nd term = a + d where a = 1, a + d = –3,
⇒ d = –4(common difference of A.P.)
⇒ an = a + (n – 1)d = 1 + (n–1)(–4) = 5 – 4n.
and Sn = n/2{a + an} = n/2{1 + 5 – 4n} = n(3 – 2n).
Illustration:
If 6 arithmetic means are inserted between 1 and 9/2, find the 4th arithmetic mean.
Solution:
Let a1, a2, a3, a4, a5, a6 be six arithmetic means
⇒ 1, a1, a2, ……, a6, 9/2 will be in A.P.
Now, 9/2 = 1 + 7d ⇒ 7/2 = 7d ⇒ d = 1/2.
Hence a4 = 1 + 4 (1/2) = 3.
Home