## Important Concepts and Formulas - Sequence and Series

raghav
2016-01-23 17:14:58
how to find the product of terms which are in ap and gp?
sam
2016-02-03 12:38:40
Opara
2015-11-28 08:31:48
Hi great one, thanks a million for the enlightenment.

Under the section "5. Some Interesting Properties to Note", I observed you wrote HP rather than GP here

Three non-zero numbers a, b and c are in HP if (a−b)/(b-c) = a/c
jiju (Junior Maths Expert, careerbless.com)
2015-12-23 10:14:47
It is HP only.

Suppose (a−b)/(b-c) = a/c
=> (a-b)c = (b-c)a
=> ac - bc = ab - ac
=> 2ac = ab + bc
=>  2/b = 1/c + 1/a  (divided all terms by abc)

i.e., 1/a , 1/b, 1/c are in AP
i.e.,  a,b,c are in HP
macoi
2015-06-18 12:35:06
mathematically speaking, the next term cannot be determined by giving only the first finite number of terms of a general sequence. explain this fact by giving examples
sani
2015-05-11 10:59:13
I hate sequence and series but after watching these formulas arrange in such an efficient manner i now i feel relaxed now i am not soooooooo much confuse so thanks.
pardeep
2015-03-29 09:09:19
354 180 64 21 10.2 ?
Dev
2015-04-06 20:23:11
354/2 + 3 =  180
180/3 + 4 = 64
64/4 + 5 = 21
21/5 + 6 = 10.2
10.2/6 + 7 = 8.7
Ankita
2015-03-25 05:49:29
What is the formula to find relation between ap hp
Apurv
2015-08-22 06:46:57
no formula

If a, b, c are in AP, 2b = a + c

Suppose a,b,c are in H.P. As reciprocals of A.P. are the terms in H.P,

$\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}\\ \Rightarrow \dfrac{2}{b}=\dfrac{a+c}{ac}\\ \Rightarrow \dfrac{b}{2}=\dfrac{ac}{a+c}\\ \Rightarrow b=\dfrac{2ac}{a+c}$
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