(iv) Given that, a3=15,S10=125As an=a+(n−1)d,a3=a+(3−1)d15=a+2d……(i)Sn=n2[2a+(n−1)d]S10=102[2a+(10−1)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=a−2a=17a10=a+(10−1)da10=17+(9)(−1)a10=17−9=8