Question

In A.P a3=15,S10=125 then find d and a10

Answer 1

Answer:

Given:

a3=15

a+(3-1)d=15

a+2d=15_(1)

S10=10/2(2a+(10-1)d)

125=5(2a+9d)

125=10a+45d_(2)

solving eq -1 and eq- 2

S10×1;10a+45d=125

a3×10;10a+20d=20

25d=105

d=105/25

{d=21/5}

substitute d value in eq- (1)

a+2(21/5)=15

a+42/5=15

5a+42=75

5a=75-42

{a=33/5}

a10=a+(10-1)d

a(10)=33/5+9(21/5)

a10=33/5+189/5

{a10=222/5}

Answer 2

Answer: d=-1, a10=8

Step-by-step explanation:

An = a+(n-1)d

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

A3=a+(3-1)d

15=a+2d (1)

S10=10/2[2a+(10-1)d]

125=5[2a+9d]

25=2a+9d (2)

Multiple equation(1)*2times

(15=a+2d)2

30=2a+4d (3)

Eq3- Eq2

2a+4d=30

2a+9d=25

Answer=-5d=5

d=-1

Substitute "d" in eq (2)

25=2a+9(-1)

25=2a-9

34=2a

a=17

Answer a10=17+(10-1)(-1)

=17-9

=8

Answers d=-1, a10=8