Question

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12

Answer 1

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Answer 2

a. = a +(n-1)d
n
so a. = a +(3-1)d
3
a = a+2d
3
16 a +2d
a+2d=16

a = a +(8-1)d
7
a = a+6d
7

a = a+(4-1)d
5
a =a+4d
5

given that
7th term exceeds the 5th term by 12
7th term-5th term =12

a - a= 12
7 5
(a+6d) -(a+4d)= 12
a+6d-a+4d=12
a-a+6d-4d =12
0+6d-4d=12
2d=12
d = 12\2
d = 6

putting the value of d in (1)

a=16-2d
=16 -2×6=16-12=4
hence
first term of ap= a=4

second term of ap=10

third term of ap = 16

so ap is 4,10,16
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