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a) 1,2,4,.... to 10 terms
Solution:
Here,
a = 1
r = \dfrac{2}{1} = 2
n = 10
Now,
Sum of geometric series (S_n) = \dfrac{a(r^n -1)}{r-1}

= \dfrac{1(2^10 -1)}{2-1}
= \dfrac{1024 -1}{1}
= 1023


c) 1,-3,9,-27,... to 9 terms
Solution:
Here,
a = 1
r = \dfrac{-3}{1} = -3
n = 9
Now,
Sum of the geometric series (S_n) = \dfrac{a(r^n -1)}{r-1}

= \dfrac{1[(-3)^9 -1]}{-3-1}

= \dfrac{-19683 - 1}{-4}
= \dfrac{-19684}{-4}
= 4921

e) 32,48,72,...to 8 terms
Solution:
Here,
a = 32
r = \dfrac{48}{32} = \dfrac{3}{2}

n = 8
Now,
Sum of the geometric series (S_n) = \dfrac{a (r^n -1)}{r-1}

= \dfrac{32 \left [ \left ( \dfrac{3}{2} \right )^8 -1 \right ]}{\dfrac{3}{2} -1}

= \dfrac{32 \left[ \dfrac{6561}{256} - 1 \right ]}{ \dfrac{3-2}{2} ]

= \dfrac{32 \left [ \dfrac{6561-256}{256} \right] }{\dfrac{1}{2}}

= 32 × \dfrac{6305}{256} × \dfrac{2}{1}

= \dfrac{32 × 6305 × 2}{256}

= \dfrac{6305}{4}
= 1576 \dfrac{1}{4}


2. Find the sum of the geometric series whose first term is 1, common ratio 3 and last term 243.

Solution:
Here,
a = 1
r = 3
b = 243
Now,
Sum of the geometric series (S_n) = \dfrac{br - a}{r-1}

= \dfrac{243×3 -1}{3-1}
= \dfrac{729-1}{2}
= \dfrac{728}{2}
= 364


3. Find the sum of the following sequence,

a) 2^1, 2^2, 2^3, 2^4, ..... , to 10 terms

Solution:
Here,
a = 2^1 = 2
r = \dfrac{2^2}{2^1} = 2
n = 10
Now,
Sum of the geometric series (S_n) = \dfrac{a(r^n -1)}{r-1}

= \dfrac{2(2^10-1)}{2-1}
= \dfrac{2×(1024-1)}{1}
= 2×1023
= 2046


4. Find the number of terms and the common ratios of a geometric series when,

a) The first term is 7, the last term 189 and the sum 280

Solution:
Here,
a = 7
b = 189
S_n = 280
We know,
S_n = \dfrac{br - a}{r-1}
or, 280 = \dfrac{189r -7}{r-1}

or, 280(r -1) = 189r-7
or, 280r - 280 = 189r -7
or, (280-189)r = 280-7
or, 91r = 273
\therefore r = 3
Now,
b = ar^{n-1}
or, 189 = 7 × 3^{n-1}
or, \dfrac{189}{7} = 27 = 3^{n-1}

or, 3^3 = 3^{n-1}
or, 3 = n-1
\therefore n = 3+1 = 4
Hence, the required number of terms and the common ratio of the given GS are 4 and 3, respectively.


5. Find the sum of the following series.

a) \sum^{5}_{n=3} 3(4)^{n-2}

Solution:
Here,
t_1 = 3×4^{3-2} = 3×4^1 = 12

t_2 = 3×4^{4-2} = 3×4^2 = 3×16 = 48

t_3 = 3×4^{5-2} = 3×4^3 = 3×64 = 192

Now,
Sum of given series (S_n) = t_1 + t_2 + t_3

= 12 + 48 + 192
= 252

6. How many terms must be taken of:

a) The series 32+48+72+,...., to make the sum 665?

Solution:
Here,
a = 32
r = \dfrac{48}{32} = \dfrac{3}{2}

S_n = 665
We know,
S_n = \dfrac{a(r^n -1)}{r-1}

or, 665 = \dfrac{32 \left [ \left ( \dfrac{3}{2} \right )^n -1\right ] }{ \dfrac{3}{2} -1 }

or, 665 × \left ( \dfrac{3}{2} -1 \right ) = 32 \left [ \left ( \dfrac{3}{2} \right ) ^n -1 \right]

or, 665 × \dfrac{3-2}{2} = 32 \left [ \left ( \dfrac{3}{2} \right )^n -1 \right ]

or, \dfrac{665}{2} × \dfrac{1}{32} = \left ( \dfrac{3}{2} \right )^n -1

or, \dfrac{665}{64} + 1= \left ( \dfrac{3}{2} \right )^n

or, \dfrac{665 + 64}{64} = \left ( \dfrac{3}{2} \right )^n

or, \dfrac{729}{64} = \left ( \dfrac{3}{2} \right )^n

or, \left ( \dfrac{3}{2} \right )^6 = \left ( \dfrac{3}{2} \right )^n

\therefore n = 6
Thus, 6 terms of the series add up to 665.


7 a) If the first term of a positive geometric series is 3 and 3rd term is 12, find the sum of its first 8 terms.

Solution:
a = 3
t_3 = 12
To find: S_8=?
We know,
t_3 = ar^2
or, 12 = 3 × r^2
or, r^2 = 4
\therefore r = \pm2
Here, r = +2 [positive geometric series]

Now,
S_8 = \dfrac{a(r^n -1)}{r-1}

= \dfrac{3(2^8 -1)}{2-1}
= \dfrac{3(256-1)}{1}
= 3×255
= 765


8 a) The sum of first 3 terms of a GS is 292 and its common ratio is 8. Find the first term of the series.

Solution:
Here,
S_3 = 292
r = 8
To find: a = ?
We know,
S_n = \dfrac{a(r^n-1)}{r-1}
or, S_3 = \dfrac{a(8^3 -1)}{8-1}

or, 292 = \dfrac{a(512-1)}{7}

or, 292 = a ×  \dfrac{511}{7}

or, a = \dfrac{292}{73}
\therefore a = 4
9 a) Sum of 1st 6 terms of a GP is 28 and sum of the first 3 terms is 1. Find its common ratio and its first term.

Solution:
Here,
S_6 = 28
S_3 = 1
We know,
S_n = \dfrac{a(r^n-1)}{r-1}
Now,
S_6 = \dfrac{a(r^6 -1)}{r-1}

or, 28 = \dfrac{a(r^6 -1)}{r-1}

or, (r-1) = \dfrac{a(r^6-1)}{28} - (i)

And,
S_3 = \dfrac{a(r^3-1)}{r-1}
or, 1 = \dfrac{a(r^3-1)}{r-1}

or, (r-1)1= (r^3-1)a
or, (r-1) = a(r^3 -1) - (ii)

Dividing equation (i) by (ii), we get,

or, \dfrac{(r-1)}{(r-1)} = \dfrac{ \dfrac{a(r^6-1)} {28} }{ \dfrac{a(r^3-1)}{1}}

or, 1 = \dfrac{[ (r^3)^2 -1^2]}{28 (r^3 -1)}

or, 1 = \dfrac{(r^3+1)(r^3 -1)}{28(r^3 -1)}

or, 1 = \dfrac{r^3+1}{28}
or, 28 = r^3 +1
or, 28-1 = r^3
or, r^3 = 27
or, r^3 = 3^3
\therefore r = 3
Put value of r in equation (ii), we get,

or, (3-1) = a(3^3-1)
or, 2 = a (26)
\therefore a = \dfrac{1}{13}