Question

In an apartment given a3=15,s10=125,find d and a10.

Answer 1

A3 = 15

=> a + 2d = 15

=> 2a +4d = 30 ------(1)

S10 = 125

=> 5[ 2a + 9d] = 125

=> 2a +9d = 25 ------(2)

On subtracting equation 1 from 2, we get

5d = - 5

d = - 1

a = 17

A10 = a + 9d

= 17 + 9 * (-1)

= 17 - 9

= 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