Question

One more question :

Find the sum of G.P

1, -a , a^2, -a^3, ....n terms (if a ≠ -1)​

Answer 1

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$


$\boxed{ \bf{ \sf{GIVEN : 1, -a , {a}^{2} , {a}^{3} \: ... \: n \: terms \: }}}$


$\boxed{ \bf{ \sf{TO \: FIND}} : \:{ \sf{ Sum \: of \: the \: G.P \: }}}$



$\boxed{ \bf{ \sf{SOLUTION : \: }}}$


$a \: = \: 1 \:, \: \: r \: = \frac{ {a}^{2} }{a} = a$


Now,


$\implies \: S_{n} \: = \: \frac{a( {r}^{n} - 1) }{r - 1}$

$\implies \: S_{n} \: = \: \frac{1( { - a}^{n} - 1) }{ - a- 1}$

$\implies \: S_{n} \: = \: \frac{ - ( { a}^{n} + \: 1) }{ - ( a \: + \: 1)}$

$\implies \: S_{n} \: = \: \frac{ ( { a}^{n} + \: 1) }{ ( a \: + \: 1)}$

Answer 2

Step-by-step explanation:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196

1 2 3 4 5 6 7 8 9 10 11 12 13 14

squares up to 200

21=t*(1 - r^3)/(1 - r)

3+6+ 12=21

9+ 36+144=189

The numbers are 3, 6 12