# AP

\(CIE\0606\2020\w\paper 12\Q4

The 7 th and 10 th terms of an arithmetic progression are 158 and 149 respectively.

(a) Find the common difference and the first term of the progression.

(b) Find the least number of terms of the progression for their sum to be negative.

$*********$ Solution $********$

Since $u_n=u_1+(n-1)d$, for an arithmetic progression,

$u_7=u_1+6d=158$    $\cdots$    (1)

$u_{10}=u_1+9d=149$  $\cdots$    (2)

$(2)-(1): 3d=-9\Rightarrow d=-3.$ Thus $u_1=158-6(-3)=176.$

$S_n=\frac{n}{2}\left[ 2(176)+(n-1)(-3)\right]<0$

Since $n>0, 352-3n+3<0\Rightarrow n>118.3.$ Thus $n=119.$

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