In a geometric progression,the third term exceeds the second by 6 and the second exceeds the first by 9.find (i) the first term (ii) the common ratio (iii) the sum of the first ten terms
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Third term = ar²
Second term = ar
First term = a
Given, ar² = ar + 6
ar(r-1) = 6 ....(1)
Also, ar = a+ 9
a(r-1) = 9 .
Substitute in (1)
9r = 6
r = 2/3 .
Substitute in any equation to get value of a.
a( 2/3 - 1 ) = 9
a ( -1/3 ) = 9
a = -27 .
Therefore, the first term is -27 . Common ratio is 2/3 .
Sum of ten terms = -27 ( (-1/3)^10 - 1 ) / -1/3 -1
= -27 [ ( 3^10- 1 )/3^10 ] * 3 / -4
= - 81 ( 59048 ) / 59049 ) / -4
= 20.24 .
Second term = ar
First term = a
Given, ar² = ar + 6
ar(r-1) = 6 ....(1)
Also, ar = a+ 9
a(r-1) = 9 .
Substitute in (1)
9r = 6
r = 2/3 .
Substitute in any equation to get value of a.
a( 2/3 - 1 ) = 9
a ( -1/3 ) = 9
a = -27 .
Therefore, the first term is -27 . Common ratio is 2/3 .
Sum of ten terms = -27 ( (-1/3)^10 - 1 ) / -1/3 -1
= -27 [ ( 3^10- 1 )/3^10 ] * 3 / -4
= - 81 ( 59048 ) / 59049 ) / -4
= 20.24 .
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