In an apartment given a3=15,s10=125,find d and a10.
Vik231:
Sorry I written apartment it's ap
Answers
Answered by
6
A3 = 15
=> a + 2d = 15
=> 2a +4d = 30 ------(1)
S10 = 125
=> 5[ 2a + 9d] = 125
=> 2a +9d = 25 ------(2)
On subtracting equation 1 from 2, we get
5d = - 5
d = - 1
a = 17
A10 = a + 9d
= 17 + 9 * (-1)
= 17 - 9
= 8
=> a + 2d = 15
=> 2a +4d = 30 ------(1)
S10 = 125
=> 5[ 2a + 9d] = 125
=> 2a +9d = 25 ------(2)
On subtracting equation 1 from 2, we get
5d = - 5
d = - 1
a = 17
A10 = a + 9d
= 17 + 9 * (-1)
= 17 - 9
= 8
Answered by
0
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
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