How to do this ? If an arithmetic progression consists of 31 terms , in which the 16th term is m,then find sum of all the terms of this ap. ( 1 mark )

Question

anonymous · Answer

$$m=a+15d \ S_{31}=\frac{31}{2}(2a+30d)=\frac{31}{2}2(a+15)=31m$$

anonymous · Answer

thanks a lot @Jonask

anonymous · Answer

there is a mistake i.e
$$\frac{ 31 }{ 2 }2(a+15d)=31(a+15d)=31m$$

anonymous · Answer

hmmmm! that's understandable @Salmon

anonymous · Answer

Ya..and I liked the way you explained it...

anonymous · Answer

Ya..and I liked the way you explained it...@Jonask

anonymous · Answer

I am doubtful. sum of terms from 1 to n,   (1 +2 + 3...), is (n/2)(n+1), so 1 to 10 sum to 55, for example. This seems at odds with solution given..

anonymous · Answer

here we are not dealing with consecutive terms so we cant use the analogy of n/2(n+1),the numbers have a difference we didnt compute,we dont need to even comute it.....suppose $$d=a=1,m=16\1+2+3+4...+31\S_{31}=16\times 31=\frac{31}{2}(31+1)$$ wich is of the form n/2(n+1)...

anonymous · Answer

An arithmetic progression increases a constant amount A between terms, thus you can turn it into a progression that looks like
A(1+2+3...n) and thus you can use the equation I gave, 
(n/2)(N+1),  multiplying what you get by A. Yes, some manipulation is required, but they ask us not to give the answers.