In an arithmetic progression, if $S_{40}=1030$ and $S_{12}=57$, then $S_{30}-S_{10}$ is equal to :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $S_{40}=20(2a+39d)=1030\Rightarrow2a+39d=51.5$; $S_{12}=6(2a+11d)=57\Rightarrow2a+11d=9.5$. Subtracting: $d=\frac32$, $a=-\frac72$. $S_{30}-S_{10}=15(2a+29d)-5(2a+9d)=15(36.5)-5(6.5)=515$. Answer (3).

Q. In an arithmetic progression, if $S_{40}=1030$ and $S_{12}=57$, then $S_{30}-S_{10}$ is equal to :

Explanation

Reviewed by Dr. Abubacker

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