Answer:
Given, \[{{a}_{10}}=52;{{a}_{17}}={{a}_{13}}+20\] \[\Rightarrow \] \[a+16d=a+12d+20\] \[\Rightarrow \] \[16d=12d+20\] \[\Rightarrow \] \[4d=20\] \[\Rightarrow \] \[d=\frac{20}{4}=5\] Also, \[a+9d=52\] \[\Rightarrow \] \[a+9\times 5=52\] \[a+45=52\] \[a=7\] Therefore A.P. = 7, 12, 17, 22, 27....
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