Question

a3=15 , s10 =125 find, d and a10

Answer 1

a3 = 15

=> a +2d = 15

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

S10 = 125

=> 10/2 [ 2 *a + (10-1) d] = 125

=>5(2a +9d) = 125

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

On subtracting equation 2 from 1, we get

-5d = 5

=> d = - 1

Now,

On putting the value of d in equation 1, we get

2a - 4 = 30

=> a = 34/2

=> a = 17

A10 = a + 9d

= 17 - 9

= 8

Answer 2

a3 = 15

a + 2d = 15 ---------- equation 1

S10 = 125

n/2 (2a + (n-1)×d) = 125

10/2 (2a + ( 10-1) × d) = 125

5 (2a + 9d) = 125

2a + 9d = 125 / 5

2a + 9d = 25 ------------- equation 2

by equation 1 × 2 and 2 × 1

we will get :

2a + 4d = 30

2a + 9d = 25. (Now subtract them by changing their sign)

- 5d = 5

d = -1 ------------ putting in equation 2

2a + 9d = 25

2a - 9 = 25

2a = 25 + 9

a = 34 / 2

a = 17 .

a10 = a + 9d

= 17 - 9

= 8.

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