Question

3. Find Sn for each of the geometric series described below.
(i) a = 3, t 384, 8 = n = 8. (ii) a = 5, r = 3, n = 12.

Answer 1

(a) a = 3,t₈ = 384,n=8
We know, tr = arⁿ⁻¹
384 = 3 × r⁸⁻¹
128 = r⁷
(2)⁷ = r⁷
r = 2
Now, sum of n terms in GP , Sn = a[rⁿ - 1]/(r - 1)
S₈ = 3[2⁸ - 1]/[2 - 1]
= 3[2⁸ - 1]


(b) a = 5, r = 3, n = 12
Sn = a[rⁿ - 1]/(r - 1)
=5[3¹² - 1]/2
Hence, sum of 12 terms = (2.5)[3¹² - 1]

Answer 2

GEOMETRIC PROGRESSION :
A sequence a1, a2, a3 ………….. an Is called a geometric sequence if a(n+1) = anr, where r is a non zero constant. Here, a1 is the first term and the constant r is called common ratio. The sequence is also called a geometric progression(G.P)

a1 = a , r = a(ⁿ+1)/ aⁿ

General term of a geometric sequence is
tn = arⁿ-1
General form of G.P is a, a r , a r ²,.........

Sum of n terms in GP ,Sn = a( rⁿ - 1 ) / ( r - 1)

GIVEN :
t₈ = 384, a = 3 , n = 8
First we have to find r.

General term of a geometric sequence is
tn = arⁿ-1
384 = 3 × r^(8-1)
3r⁷ = 384
r⁷ = 384/3
r⁷ = 128
r⁷ = 2⁷
r = 2
Sum of n terms in GP ,Sn = a( rⁿ - 1 ) / ( r - 1)
S₈ = 3 (2⁸ - 1)/(2-1)
S₈ = 3 (256 - 1)/(1)
S₈ = 3 (255)
S₈ = 765

Hence, S8 = 765

(ii) GIVEN :
a = 5,r = 3,n=12
sn = a( rⁿ - 1 ) / ( r - 1)
S₁₂ = 5 (3¹² - 1) / (3-1)
S₁₂ = 5 (3¹² - 1) / (2)
S₁₂ = (5/2) (3¹² - 1)
S₁₂ = (5/2) (3¹² - 1)

Hence, S12 = (5/2) (3¹² - 1)

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