Question

Find the indicated term of each geometric Progression
(i) a1 = 9; r = 1/3; find a7 (ii) a1 = −12; r = 1/3; find a6

Answer 1

if first term, $a_1$ and common ratio, r is given of geometric progression then,

$a_n=a_1r^{n-1}$

(i) $a_1=9$ and r = 1/3

$a_7=a_1r^{7-1}$

$\implies a_7=9(1/3)^6=\frac{1}{81}$

$\implies a_7=\frac{1}{81}$

(ii) $a_1=-12$ and r = 1/3

$a_6=a_1r^{6-1}$

$\implies a_6=(-12)(1/3)^5=\frac{-12}{243}$

$\implies a_6=\frac{-4}{81}$

Answer 2

Hi ,

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Let a , r are first term and common

ratio of a G.P

n th term = an = ar^n-1

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Now ,

i ) a = a1 = 9 , r = 1/3 , n = 7

a7 = ar^6

a7 = 9 × ( 1/3 )^6

= 3²/3^6

= 1/3⁴

a7 = 1/81

ii ) a = a1 = -12 , r = 1/3 , n = 6

a6 = ar^5

a6 = ( -12 )( 1/3 )^5

a6 = ( - 4 × 3 )/( 3 × 81 )

a6 = ( -4 )/81

I hope this helps you.

: )