StudentAidUnit 6: Sequences and Series — Exercise 6 7
11th Class Mathematics · Unit 6: Sequences and Series
Q.1.Find the sum of first 15 terms of the G.P. 1, 1/3, 1/9, ...
Given
First term
a1 = 1
Common ratio
r = 13
Number of terms
n = 15
Condition
|r| = 13 < 1
Formula
Sum formula for G.P.
Sn = a1(1-rn)1-r
Working
Substituting values
S15 = 1 · (1 - (1/3)15)1 - 1/3
Simplifying
= frac{315 - 1}{315 · 32 = frac{3(315 - 1)}{2 · 315
Computing 3^15 and substituting
= 3 × 143489062 × 14348907 = 4304671828697814
Result
Sum of first 15 terms
S15 = 71744534782969
Q.2.The 3rd term of a G.P. is 16 and the 6th term is -128. Find the first term and the sum of the first seven terms.
Given
3rd term
a3 = 16
6th term
a6 = -128
Formula
G.P. term formula
an = a1 rn-1
Working
Setting up equations
a1 r2 = 16 quad (i)
And
a1 r5 = -128 quad (ii)
Dividing (ii) by (i)
r3 = -12816 = -8
Result
Common ratio
r = -2
Substituting r into (i)
a1(-2)2 = 16
Solving for a_1
4a1 = 16
First term
a1 = 4
Formula
Sum formula for G.P.
Sn = a1(rn - 1)r - 1
Working
For n = 7
S7 = 4((-2)7 - 1)-2 - 1
Computing
= 4(-128 - 1)-3 = 4 × (-129)-3
Result
Sum of first seven terms
S7 = 172
Q.5.Sum the series 2 + (1-i) + (1/i) + ... to 8 terms.
Given
First term
a1 = 2
Second term
a2 = 1 - i
Number of terms
n = 8
Common ratio
r = 1-i2
Series type
Geometric series
Formula
Sum formula
Sn = a1(rn - 1)r - 1
Computing r^8
(1-i)2 = -2i, quad (1-i)4 = (-2i)2 = -4, quad (1-i)8 = (-4)2 = 16
So
r8 = left(1-i2right)8 = 16256 = 116
And
r - 1 = 1-i2 - 1 = -1-i2
Working
Substituting
S8 = 2(1/16 - 1)(1-i)/2 - 1 = 2(-15/16)(-1-i)/2
Simplifying
= 60/161+i = 60(1-i)16(1-i2) = 60(1-i)32
Result
Sum to 8 terms
S8 = 158(1-i)