Question

a3=15,s10=125,find d and a10

Answer 1

a3 = 15
or
a+2d=15_________eq1
S10=125
or
s10=n÷2{2a+(n-1)d}
125=10÷2{2a+9d}
25=2a+9d__________eq2
multiply eq1 by 2
(a+2d=15)×2
2a+4d=30______eq3
subtract eq 3 from eq 2
2a+9d-2a-4d=25-30
5d=-5
or
d=-1
put d=-1 in eq 1
a+2d=15
a=17
a10=a+9d
a10= 17-9
a10=8

Answer 2

Answer:

an=a+(n-1)d

a3=a+(3-1)d

15=a+2d

a+2d=15 _________ {1}

Sn=n/2(2a+{n-1}d)

S10=10/2(2a+{10-1}d)

125=5(2a+9d)

125/5=2a+9d

25=2a+9d ___________{2}

solving eq{1} & eq{2}

putting a value in eq {2}

2(15-2d)+9d=25

30-4d+9d=25

5d=25-30

=-5

=d=-1

=>an=a+(n-1)d

a10=a+(10-1)d

a10=a+9d

a10=17+9(-1)

a10=17-9

a10=8