Question

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

Answer 1

Solution :

Let a and d are first term and common

difference of an A.P.

a3 = 15

=> a + 2d = 15

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

S10 = 125

=> ( 10/2 )[ 2a + 9d ] = 125

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

=> 2a + 9d = 125/5

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

Substitute a = 15 - 2d in equation ( 2 ),

we get

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

=> 30 - 4d + 9d = 25

=> 5d = 25 - 30

=> 5d = -5

=> d = -5/5

=> d = -1

Therefore ,

common difference = d = -1

••••

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