Ann's session is expressed as an AP;
10,12,14...50
Joy's; 15,20,...
a=a+7d compared to Joy's
Ann Joy
a+7d+(n-1)d=a+(n-1)d
10+7(2)+2n-2=15+5n-5
10+14-2+2n=10+5n
22+2n=10+5n
12=3n
12/3=n
4=n

For Joy
a=15
d=5
n=?
After how many sessions does Joy swim the same number of lengths as ann...

@ 8th sessions
a= 10
n= 8
d= 2
Un = a+(n-1)d
Un= 10+(8-1)2
Un= 24..
Ann.....24. Joy......15
Ann.....26. Joy.......20
Ann.....28. Joy......25
Ann.....30. Joy......30

30=15+5n-5
30=5n+10
5n=20
n=4

The number of Anna's lengths=50
Therefore,we are looking for the number of sessions at which Joy's lengths will be equal to Anna's lengths(50). This suggests that our Un is 50.
And for Joy,
Un =a + (n-1) d
Where: a= 15, n=?,d= 5 and Un =50
Substite the values
50= 15 + (n-1) 5
50=15 + 5n - 5
50= 15-5 + 5n
50= 10 + 5n
Collect like terms
50-10=5n
40=5n
n=40/5
n=8.
Therefore,after 8 sessions, Joy would have swum the same number of lengths as Ann( 50 lengths)

Assuming the name Sue was a typographical error for Ann,

Firstly, we solve for the lengths that Ann is currently swimming by her 8th session,
\( A_{8} \)=10 + (8-1)2
\( A_{8} \)=10 + 14 = 24.
Therefore, we know that Ann is swimming 24lengths by the time Joy joins her.

Now to solve for how many sessions Joy will swim before she starts swimming on par with Ann, let's make 24 the first term for Ann,\( a \)
\( J_{n}= 15 + (n-1)5 \)
\( A_{n}= 24 + (n-1)2 \)
Thus,
\( J_{n}= 15 + (n-1)5 \)=\( A_{n}= 24 + (n-1)2 \)
15 + (n - 1)5 = 24 + (n - 1) 2
15 + 5n - 5 = 24 + 2n - 2
10 + 5n = 22 + 2n
5n - 2n = 22 - 10
3n = 12
n = 4.

Therefore, Joy has to swim 4 sessions before she swims the same number of lengths as Ann.