Question

in a Ap given a3=15,s10=125 find d and a10

Answer 1

S10=10÷2 (2a +(10-1)d=5 (2a+9d)
125=5 (2a+9d)
125÷5=2a+9d


25=2a+9d _(1)
a3=a+2d
15=a+2d _(2)
subtracting equation (2) from (1) ,we get
2a+9d -a -2d =25-15
a+7d=10
a=10-7d _(3)
put the value of a in equation (2)
15=10-7d +2d
d=-1
put d=-1 in (3)
a=10-(-1×7)
a=17
now, a10=a+9d
a10=17-9
a10=8

a=

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