Now, let us try to find out the sum of the first n terms of a G.P.
Sn = a + ar + ar2 +…+ arn–1 … (6)
Multiplying both sides by r, we get,
rSn = ar + ar2 +…+ arn–1 + arn … (7)
Subtracting (7) from (6), we have
Sn – rSn = a – arn
or, Sn = a(1–rr)/(1–r) … (8)
From equation (8) sum of n terms of a G.P.
Sn = a(1–rr)/(1–r)
Basically we have to find out the value of particular case when, n tends to infinity.
If we take any value of |r| greater than 1 then value of rn when n → ∞ will tend to infinite. Hence value of S∞ will also tend to infinite. If we take of |r| less than rn when n → ∞ will tend to zero. (take any number greater than 1 multiply it by itself several time. Are you getting convinced that it will tend to infinity. Do the same exercise with one another number less than 1).
Thus S∞ = a/1–r for |r| < 1 … (9)
Enquiry : What happens when any real number can be add subtract, multiply and divides to each term of a geometric series?
1. Multiplication/Division by a constant number to each term of a G.P. also results a G.P.
Suppose a1, a2, a3, ……, an are in G.P.
then ka1, ka2, ka3, ……, kan and
a1/k, a2/k, ... ... ... an/k will also be in G.P.
Where k ? R and k ≠ 0.
2. Multiplication/Division of two G.P.’s also results a G.P.
Suppose a1, a2, a3, ……, an
and b1, b2, b3, ……, bn are two G.P.
then a1b1, a2b2, a3b3, ……, anbn
then a1/b2, a2/b2, ... ... ..., an/bn will also be in G.P.
3. Reversing the order of a G.P.’s also results a G.P.
Suppose a1, a2, a3, ……, an are in G.P.
then an, an–1, an–1, ……, a3, a2, a1 will also be in G.P.
4. Taking the inverse of a G.P. also results a G.P.
Suppose a1, a2, a3, ……, an are in G.P.
then 1/a1, 1/a2, 1/a ……, 1/an will also be in G.P.
Note:
Students are suggested to assume the known variable related to geometric progression in following way.
Three number in G.P. ∴ α/ß, α, αß c.r = ß
Four number in G.P. ∴ α/ß3, α ß, αß, αß3 c.r. = ß2
Five numbers in G.P. ∴ α . αß . αß2 – c.r. = ß
Summary of Important Notes:
• If each term of a G.P. is multiplied (or divided) by a fixed non-zero constant, then the resulting sequence is also a G.P. with same ratio as that of the given G.P.
• If each term of a G.P. (with common ratio r) is raised to the power k, then the resulting sequence is also a G.P. with common ratio rk.
• If a1, a2, a3, ……, b1, b2, b3, …… are two G.P.’s with common ratios r and r’ respectively then the sequence a1b1, a2b2, a3b3, …… is also a G.P. with common ratio rr’.
• If we have to take three terms in a G.P., it is convenient to take them as a/r, a, ar. In general, we take a/rk, a/rk–1, …, a, ar, …, ark in case we have to take (2k + 1) terms in a G.P.
• If we have to take four terms in a G.P., it is convenient to take them as a/r3, a/r, ar, ar3. In general, we take a/r2k–1, a/r2k–3, ... ... ... a/r, ar, ……, ar2k–1, in case we have to take 2k terms in a G.P.
• If a1, a2, ……, an are in G.P., then a1an = a2an–1 = a3an–2 = ……
• If a1, a2, a3, …… is a G.P. (each a1 > 0), then loga1, loga2, loga3 …… is an A.P. The converse is also true.
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