## 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?

please tell me the formula

please tell me the formula

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

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

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.

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 =

180/3 + 4 = 64

64/4 + 5 = 21

21/5 + 6 = 10.2

10.2/6 + 7 =

**8.7**
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}@@

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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