A student read common difference of an A.P as -3 instead of 3 and obtained the sum of first 10 terms -30. Then, the actual sum of first 10 terms is equal to

Let common difference d1=−3  and first term be a . ∴    Series become a,  a−3,  a−6,  ...  ,a−27 ∴       S=10a+(−3−6−  ...  −27) ⇒   −30=10a−3(1+2+  ....  +9) ⇒    −30=10a−3[9(9+1)2] ⇒    −30=10a−135 ⇒     10a=105 ⇒    a=10510 Now,  correct common difference d2=3 ∴   Required sum=102[2×10510+(10−1)3]                                   =5[1055+27]                                =5×48=240