in an A.P. GIVEN a3=15 ,S10=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
Answer:
Given that, a3 = 15, S10 = 125
As we know, from the formula of the nth term in an AP,
an = a +(n−1)d,
Therefore, putting the given values, we get,
a3 = a+(3−1)d
15 = a+2d ………………………….. (i)
Sum of the nth term,
Sn = n/2 [2a+(n-1)d]
S10 = 10/2 [2a+(10-1)d]
125 = 5(2a+9d)
25 = 2a+9d ……………………….. (ii)
On multiplying equation (i) by (ii), we will get;
30 = 2a+4d ………………………………. (iii)
By subtracting equation (iii) from (ii), we get,
−5 = 5d
d = −1
From equation (i),
15 = a+2(−1)
15 = a−2
a = 17 = First term
a10 = a+(10−1)d
a10 = 17+(9)(−1)
a10 = 17−9
= 8