(iv) given a3= 15. S. = 125, find d and a10
Answers
Answer:
d = -1 and a10 = 8
Step-by-step explanation:
Formula:-
nth term of AP, tn = a + (n-1) d
Sum of n terms Sn = n/2[2a + (n-1)d]
a - first term and d = common difference
To find a and d
It is given that, a3=15 ,S10=125
We can write a + 2d = 15 ----(1)
10/2[2a + 9d ] = 125
⇒5[2a + 9d ] = 125
⇒2a + 9d = 25 ----(2)
(1)*2 ⇒ 2a + 4d = 30 ---(3)
(2) - (3) ⇒
5d = -5
d = -1
eq (1) ⇒ a + 2d = 15
a + -1*2 = 15
a = 15 + 2 = 17
To find a10
a10 = a + 9d = 17 + 9*-1 = 17 - 9 = 8
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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