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We will learn how to find the sum of n terms of the Geometric Progression {a, ar, ar2, ar3, ar4, ...........}
To prove that the sum of first n terms of the Geometric Progression whose first term βaβ and common ratio βrβ is given by
Sn = a(rnβ1rβ1)
β Sn = a(1βrn1βr), r β 1.
Let Sn denote the sum of n terms of the Geometric Progression {a, ar, ar2, ar3, ar4, ...........} with first term βaβ and common ratio r. Then,
Now, the nth terms of the given Geometric Progression = a β rnβ1.
Therefore, Sn = a + ar + ar2 + ar3 + ar4 + ............... + arnβ2 + arnβ1 ............ (i)
Multiplying both sides by r, we get,
rSn = ar + ar2 + ar3 + ar4 + ar4 + ................ + arnβ1 + arn ............ (ii)
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On subtracting (ii) from (i), we get
Sn - rSn = a - arn
β Sn(1 - r) = a(1 - rn)
β Sn = a(1βrn)(1βr)
β Sn = a(rnβ1)(rβ1)
Hence, Sn = a(1βrn)(1βr) or, Sn = a(rnβ1)(rβ1)
Notes:
(i) The above formulas do not hold for r = 1. For r = 1, the sum of n terms of the Geometric Progression is Sn = na.
(ii)When the numerical value of r is less than 1 (i.e., - 1 < r < 1), then the formula Sn = a(1βrn)(1βr) is used.
(iii) When the numerical value of r is greater than 1 (i.e., r > 1 or, r < -1) then the formula Sn = a(rnβ1)(rβ1) is used.
(iv) When r = 1, then Sn = a + a + a + a + a + .................... to n terms = na.
(v) If l is the last term of the Geometric Progression, then l = arnβ1.
Therefore, Sn = a(1βrn1βr) = (aβarn1βr) = aβ(arnβ1)r(1βr) = aβlr1βr
Thus, Sn = aβlr1βr
Or, Sn = lrβarβ1, r β 1.
Solved examples to find the Sum of first n terms of the Geometric Progression:
1. Find the sum of the geometric series:
4 - 12 + 36 - 108 + .............. to 10 terms
Solution:
The first term of the given Geometric Progression = a = 4 and its common ratio = r = β124 = -3.
Therefore, the sum of the first 10 terms of the geometric series
= a β rnβ1rβ1, [Using the formula Sn = a(rnβ1)(rβ1) since, r = - 3 i.e., r < -1]
= 4 β (β3)10β1β3β1
= 4 β (β3)10β1β4
= - (-3)10 - 1
= -59048
2. Find the sum of the geometric series:
1 + 12 + 14 + 18 + 116 + .............. to 10 terms
Solution:
The first term of the given Geometric Progression = a = 1 and its common ratio = r = 121 = 12
Therefore, the sum of the first 10 terms of the geometric series
S10 = a(1βr10)(1βr)
β S10 = 1 β (1β(12)10)(1β12)
β S10 = 2(1210)
β S10 = 2(210β1210)
β S10 = 2(1024β11024)
β S10 = 1024β1512
β S10 = 1023512
Note that we have used formula Sn = a((1βrn)(1βr) since r = 1/4 i.e., r < 1]
3. Find the sum of 12 terms of the Geometric Progression 3, 12, 48, 192, 768, ................
Solution:
The first term of the given Geometric Progression = a = 3 and its common ratio = r = 123 = 4
Therefore, the sum of the first 12 terms of the geometric series
Therefore, S12 = ar12β1rβ1
= 3(412β14β1)
= 3(16777216β13)
= 16777216 - 1
= 16777215
4. Find the sum to n terms: 5 + 55 + 555 + 5555 + .............
Solution:
We have 5 + 55 + 555 + 5555 + ............. to n terms
= 5[1 + 11 + 111 + 1111 + .............. + to n terms]
= 59[9 + 99 + 999 + 9999 + ................ + to n terms]
= 59[(10 β 1) + (102 - 1) + (103 - 1) + (104 - 1) + .............. + (10n - 1)]
= 59[(10 + 102 + 103 + 104 + ................ + 10n) β ( 1 + 1 + 1 + 1 + ................ + 1)] n times
= 59[10 Γ (10nβ1)(10β1) β n]
= 59[109(10n β 1) β n]
= 581[10n+1 β 10 β 9n]
β Geometric Progression
11 and 12 Grade Math
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