Question

The numbers are in AP
[
begin{aligned}
a &=10, d=-1 / 3
therefore P_{n} &=frac{n}{2}[2 a+(n-1) d]=155
end{aligned}
]
t) ( nleft[20+(n-1)left(-frac{1}{3}right)right]=310 )
( 20 n+-frac{n^{2}}{3}+frac{n}{3}=310 )
( Rightarrow frac{n^{2}}{3}+frac{61}{3} n+310=0 )
A) ( n^{2}-61 n+930=0 )
( x=3 )
( A_{10}=15[20+20(-1 / 3)]=operatorname{Ren}[=200 )
( D_{51}=frac{31}{2}[20+3 )
( P_{30}=15[20+29(-1 / 3)=155 )
( S_{31}=frac{31}{2}left[20+left(30left(frac{-1}{3}right)right)right]=155 )
( h=30 ) an 31 2

# g & Statistics - For CA Foundation series 58. The number of the the of the number terms of 10+9+92 +9.... will amount to 155 is 3 (a) 30 (b) 31 (c) 32 (d) None of these in numbers of terms of an AP series

Solution