Question

a3 =15 S10= 125 find d and a10​

Answer 1

Answer:

d = -1 and a10 = 8

Step-by-step explanation:

Formula:-

nth term of AP, tn = a + (n-1) d  

Sum of n terms Sn = n/2[2a + (n-1)d]

a - first term and d = common difference

To find a and d

It is given that, a3=15 ,S10=125

We can write a + 2d = 15  ----(1)

10/2[2a + 9d ] = 125

⇒5[2a + 9d ] = 125

⇒2a + 9d = 25  ----(2)

(1)*2 ⇒ 2a + 4d = 30 ---(3)

(2) - (3) ⇒

5d = -5

d = -1

eq (1) ⇒ a + 2d = 15

a + -1*2 = 15

a = 15 + 2 = 17

To find a10

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