In class 10th of arithmetic progressions who i solve this question in easy way of exercise 5.3 of the question is In an AP(iv) Given a3 = 15, S10 = 125, find d and a10.
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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