Question

a3=15 s10=125 d= ? A10=?

Answer 1

a3 = 15
S10 = 125

a3 = a+2d
15 = a+2d
2a = 30-4d

S10 = 10/2 (2a+9d)
125 = 10a+45d
25 = 2a+9d
2a = 25-9d

30-4d = 25-9d
5 = -5d
d = -1

a3 = a+2d
15 = a+ {2×(-1)}
a = 15+2
a = 17

a10 :
= a+9d
= 17+{19×(-1)}
= 17 + (-19)
= -2

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