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If a is the first term and r  is the common ratio of the geometric progression,

then its n^th term is given by a_n = ar^n-1 

The sum S_n of the first n terms of the G.P. is given by

S_n = a (rⁿ – 1)/ (r-1), when r ≠1; = na  if r =1             

If -1 < x < 1, then limxⁿ = 0, as n →∞. Hence, the sum of an infinite G.P. is 1+x+x²+ ….. = 1/(1-x) 

If -1 < r< 1, then the sum of the infinite G.P. is a +ar+ ar²+ ….. = a/(1-r)

If each term of the G.P is multiplied or divided by a non-zero fixed constant, the resulting sequence is again a G.P.

 

If a₁, a₂, a₃, …. andb₁, b₂, b₃, … are two geometric progressions, then a₁b₁, a₂b₂, a₃b₃, …… is also a geometric progression and a₁/b₁, a₂/b₂, ... ... ..., a_n/b_n will also be in G.P.  

Suppose a₁, a₂, a₃, ……,a_n are in G.P. then a_n, a_n–1, a_n–2, ……, a₃, a₂, a₁ will also be in G.P.

Taking the inverse of a G.P. also results a G.P. Suppose a₁, a₂, a₃, ……,a_n are in G.P then 1/a₁, 1/a₂, 1/a₃ ……, 1/a_n will also be in G.P   

If we need to assume three numbers in G.P. then they should be assumed as   a/b, a, ab  (here common ratio is b)

Four numbers in G.P. should be assumed as a/b³, a/b, ab, ab³ (here common ratio is b²)

Five numbers in G.P. a/b², a/b, a, ab, ab²  (here common ratio is b)    

If a₁, a₂, a₃,… ,a_n is a G.P (a_i> 0 ∀i), then log a₁, log a₂, log a₃, ……, log a_n is an A.P. In this case, the converse of the statement also holds good.

If three terms are in G.P., then the middle term is called the geometric mean (G.M.) between the two. So if a, b, c are in G.P., then b

 = √ac is the geometric mean of a and c.

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• Let a, b and c form an H.P. Then 1/a, 1/b and 1/c form an A.P.

• If a, b and c are in H.P. then 2/b = 1/a + 1/c,  which can be simplified as b = 2ac/(a+c) 

• If a  and b are two non-zero numbers then the sequence a, H, b is a H.P.

•     The n numbers H1, H2, ……,Hn are said to be harmonic means between a and b,
• if a, H1, H2 ……, Hn, b are in H.P. i.e. if 1/a, 1/H1, 1/H2, ..., 1/Hn, 1/b are in A.P.
• Let d be the common difference of the A.P., Then 1/b = 1/a + (n+1) d ⇒ d = a–b/(n+1)ab.

• If a  andb  are two positive real numbers then A.M x H.M = G.M2

• The relation between the three means is defined as A.M  >  G.M  >  H.M

• If we need to find three numbers in a H.P. then they should be assumed as 1/a–d, 1/a, 1/a+d

• Four convenient numbers in H.P. are 1/a–3d, 1/a–d, 1/a+d, 1/a+3d 

• Five convenient numbers in H.P. are 1/a–2d, 1/a–d, 1/a, 1/a+d, 1/a+2d

 

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Geometric Mean (GM)

 

The geometric mean is similar to the arithmetic mean, but it uses the property of multiplication, to find the mean between any two numbers.

GM of two numbers a and b may be defined as G = √(ab)

For example, GM of 2 and 3 is √6.

We can find a general formula for inserting a GM between any two arbitrary numbers a and b.

 

 If G is the geometric mean between a and b, then by definition of GP a,  G,  b are in GP.

Or G2=a.b

 

We can also find a formula for inserting n GM in between two numbers a and b.

Let  G1,  G2,  G3  ….  Gn be then GMs between a and b.

Now a,   G1,  G2, G3…. Gn,  b are in GP.

By definition of the last term of a GP  b = (n+2)th term of GP.

If R=common ratio, then by definition of GP,  b = a.Rn+1      

Now G1=a?R

       G2=a?R2

         G3=a?R3

   and Gn=a?Rn

 

We will take a few examples of GM too.

 

 

Find the GM of the following and insert n geometric means between them.

 

    e and 1
    7 and tan x
    log x and sec-1x

 

Solution:

(i) The GM between e and 1 is e. To insert n GMs, we use the definition of geometric progressions.

Let  G1,   G2,  G3,  …  Gn,  be n  geometric means inserted between e and 1.

Also, let common ratio be R.

 By the definition of GP    = (n+2)th term of the GP.

 1 =eRn+1

Or R= (1/e)1/(n+1)

Hence   G1 = e⋅R = e⋅(1/e)1/(n+1)

            G2=e⋅R2=e⋅(1/e)2/(n+1)

            G3=e⋅R=e⋅(1/e)3/(n+1)

            and  Gn=e⋅R=e⋅(1/e)n/(n+1)

 

 

(ii)   Proceeding as before the GM  of 7 and tanx is √(7tanx).

Let G1, G2, G3,… Gn,  be n  geometric means inserted between 7 and tanx.Also, let common ratio be R.

To get the n geometric means, we proceed as before.

Here tanx=(n+2)th term.

 tanx=7⋅Rn+1

Or Rn+1= tanx/7

R = (tanx/7)1/n+1

 G1 = 7R = 7⋅(tanx/7)1/n+1

    G2  =  7R2   = 7⋅(tanx/7)2/n+1

    G3 = 7R3 = 7⋅(tanx/7)3/n+1

 

and   Gn=7Rn=7(tanx/7)n/n+1

 

(iii)  Proceeding exactly as before, the GM of logx and sec-1x will be √(logx⋅sec-1x)

 Let   G1,  G2,  G3, …  Gn,  be n  geometric means inserted between logx and sec-1x.

Also, let common ratio be R.

    

To insert n GMs between logx and sec-1x, we take the sequence as

    logx, G1, G2, G3,… Gn, sec-1x.

 Gn = (n+2)th term =  sec-1x = logx.Rn+1

 R= (sec-1x/logx)1/(n+1)

Now as before G1=logx⋅R=logx⋅(sec-1x/logx)1/(n+1)

                       G2=logx⋅R2=logx⋅(sec-1x/logx)2/(n+1)

                       G3=logx⋅R3=logx⋅(sec-1x/logx)3/(n+1)

           and Gn=logx⋅Rn=logx⋅(sec-1x/logx)n/(n+1)

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Let the first term be a and the common difference be b. So, the 10th term: t10 = a + 9b = 5. and, the 18th term: t18 = a + 17b = 77. So, from t17 - t10 we get: 8b = 77 - 5 = 72. This gives: b = 9. From t10 we get: a = 5 - 9b = 5 - 9x9 = -76. So, a = -76, b = 9 and the A.P is:- -76, -67, -58, -49, ..... (Answer). read less

Let the first term be (a - 5b) and the common difference be 2b. So the six terms are: (a - 5b), (a - 3b), (a - b), (a + b), (a + 3b), (a + 5b). By the given conditions:- Sum of six terms=(a - 5b) + (a - 3b) + (a - b) + (a + b) + (a + 3b) + (a + 5b) = 6a=0. So, a=0. The fourth term=a + b=2 --> 0 + b=2 --> b = 2. So, for the A.P. the first term x = (a - 5b) = -10. The common difference, y = 2b = 4. Sum of 1st 30 terms, S30 = (n/2)[2x + (n-1)y] = 15 x [ -20 + 29 x 4] = 1440. (Answer) read less

Let the 3 terms are (a - b), a, and (a + b). So, (a - b) + a + ( a + b) = 27=> 3a=27 => a=9. Now the product is 648. So, (9 - b) x 9 x (9 + b) = 648=> (81 - b^2) = 648/9 = 72=> b^2 = 81 - 72=9 => b = =3 or b = -3. For, b = 3, the numbers are: (9 - 3), 9, (9 + 3) = 6, 9, 12 or (9 - (-3)), 9 , (9 + (-3)) = 12, 9, 6 The three numbers are 6, 9, 12 or 12, 9, 6 (Answer) read less

Let the first term and the common difference are a and b. So, the third term: t3=a + 2b=4. and, the 9th term: t9=a + 8b=-8. So, t9 - t6=6b = -8 - 4=-12 --> b=- 2. So, from the expression of t3, we get: a=4 - 2b=4 + 4=8. Let us consider that the nth term is zero. So: tn=a + (n-1)b = 0 --> 8 + (n-1)(-2) = 0 --> 8 -2n +2 = 0 --> 2n=10 --> n = 5. Answer: The 5th term of the series is zero. read less