Question

given a3=15,S10=125,find d and a10

Answer 1

a3=a+2d=15 eq (1)
s10=n/2 (2a+(n-1)d)
s10=5 (2a+9d)
10a+45d=125 eq (2)
by solving the eq 1 n 2
10a+45d=125
a+2d=15×10
10a+20d=150

10a+45d=145
10a+20d=150 a +2d=15
25d=-5 a +2 ×-5/25=15
d=-5/25 a=115
a10 = a+9d
= 115+-5/25
=

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