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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,

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