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Question

Question 3 (iv)
In an AP:
(iv) Given a3=15,S10=125, find d and a10.

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Solution

(iv) Given that, a3=15,S10=125As an=a+(n1)d,a3=a+(31)d15=a+2d(i)Sn=n2[2a+(n1)d]S10=102[2a+(101)d]125=5(2a+9d)25=2a+9d(ii)On multiplying equation (i) by 2, we get,30=2a+4d(iii)On subtracting equation (iii) from (ii), we get,5=5dd=1From equation (i),15=a+2(1)15=a2a=17a10=a+(101)da10=17+(9)(1)a10=179=8

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