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Answer 1 · 2017-12-17T03:34:40

10 years

for company A:

24000, 27200, 30400......

Arithmetic progression with $d = 3200$ $d=3200$

$a_{n} = 24000 + (n - 1) \cdot 3200$ $a_n=24000+(n-1)*3200$
$a_{n} = 20800 + 3200 n$ $a_n=20800+3200n$

For company B:

32000, 34400, 36800......

Arithmetic progression with $d = 2400$ $d=2400$

$a_{n} = 32000 + (n - 1) \cdot 2400$ $a_n=32000+(n-1)*2400$
$a_{n} = 29600 + 2400 n$ $a_n=29600+2400n$

Equate the two $a_{n}$ $a_n$ s

$a_{n} = a_{n}$ $a_n=a_n$

$20800 + 3200 n = 29600 + 2400 n$ $20800+3200n=29600+2400n$

$3200 n - 2400 n = 29600 - 20800$ $3200n-2400n=29600-20800$

$800 n = 8800$ $800n=8800$
$n = \frac{8800}{800} = 11$ $n=8800/800=11$

$n = 11$ $n=11$

the salary will be equal on the 11th term which is 10 years from the starting salary

God bless..... I hope the explanation is useful.