Question

The 5th and 13 term of an ap are 5 and -3 respectively. find the ap

Answer 1

a5=5
a13=-3

a5 = a+4d = 5
a13 = a+12d =-3

by elimination method
d=-1
and
a=9

therefore , ap= 9,8,7,6,5.....

Answer 2

tn=a+(n-1)d
t5=a+(5-1)d
5=a+4d..........1
t13=a+(13-1)d
-3=a+12d.........2
subtract 1 from 2
-3=a+12d
-
5=a+4d
- - -
_________
-8=8d
-8
____= d
8
-1=d
substituting d=-1 in 1
5=a-4
5+4=a
9=a
thus ap= 9,8,7,6,5,4,3............
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