Question

{A+2d=15} x {2a+9d=25}. =?

Answer 1

Answer:

Let a is the first term and d is the common difference of the AP

Given, a3 = 15

=> a + 2d = 15

=> a = 15 - 2d ..............1

Again given

S10 = 125

=> (10/2)*{2a + (10 - 1)d} = 125

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

=> 2a + 9d = 125/5

=> 2a + 9d = 25

From equation 1, we get

2(15 - 2d) + 9d = 25

=> 30 - 4d + 9d = 25

=> 30 + 5d = 25

=> 5d = 25 - 30

=> 5d = -5

=> d = -5/5

=> d = -1

From equation 1,

a = 15 - 2(-1)

=> a = 15 + 2

=> a = 17

Now, a10 = a + 9d

=> a10 = 17 + 9*(-1)

=> a10 = 17 - 9

=> a10 = 8

So, the value of d is -1 and a10 is 8

Answer 2

Answer:

given, find d and a10

a3=15

s10=125

from a3,

a3=15

a+2d=15

a=15-2d ------------- (1)

from s10,

s10=125

125=10/5[2a+(n-1)d]

125=5[2(15-2d)+(10-1)d]

125/5=30-4d+9d

25=30+5d

25-30=5d

-5=5d

d=-1 (ANS)

now, we will find a10

a=17, d=-1, n=10, an=??? (find)

an=a+(n-1)d

an=17+(10-1)x(-1)

an=17+9x(-1)

an=17-9

an=8 (ANS)