Arithmetic sequence nth theorem formula
an=a1+(n−1)da6=24=a1+(6−1)d ⟹ a1+5d=24......(1)a11=39=a11+(11−1)d ⟹ a1+10d=39......(2)Adding 1&2d=3a1=9So,a17=9+(17−1)∗3=57a_n=a_1+(n-1)d\\ a_6=24=a_1+(6-1)d \implies a_1+5d=24......(1)\\ a_{11}=39=a_{11}+(11-1)d \implies a_1+10d=39......(2)\\ Adding \space 1 \& 2\\ d= 3\\ a_1=9\\ So, a_{17}=9+(17-1)*3=57an=a1+(n−1)da6=24=a1+(6−1)d⟹a1+5d=24......(1)a11=39=a11+(11−1)d⟹a1+10d=39......(2)Adding 1&2d=3a1=9So,a17=9+(17−1)∗3=57
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