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Worked solution

Recover an AP and find its first sum above a threshold

The ninth term of an AP is 4343 and its first 1515 terms sum to 570570. (i) Find its first term aa and common difference dd. (ii) Given Sn>2265S_n>2265, find the least possible nn.

CJC · Sequences and Series · Self-Practice 6

(i) First term and common difference

Answer

a=3,d=5a=3,d=5.

Turn both conditions into equations

a+8d=43,152(2a+14d)=570⟹a+7d=38a+8d=43,\qquad\frac{15}{2}(2a+14d)=570\quad\Longrightarrow\quad a+7d=38

The source prints the answers; this derivation is authored working.

Subtract to eliminate the first term

d=43−38=5,a=38−7(5)=3d=43-38=5,\qquad a=38-7(5)=3

The two equations differ by one common difference.

How did this part go?

(ii) A strict sum threshold

Answer

n=31n=31.

Use the recovered progression

Sn=n2[2(3)+(n−1)5]=n(5n+1)2>2265\begin{aligned}&S_n\\ & =\frac n2[2(3)+(n-1)5]\\ & =\frac{n(5n+1)}2>2265\end{aligned}

Authored working keeps the strict inequality.

Find the first admissible integer

Sn−2265=(n−30)(5n+151)2>0⟹n>30,S30=2265,S31=2418.\begin{aligned}S_n-2265&=\frac{(n-30)(5n+151)}2>0\quad\Longrightarrow\quad n>30,\\S_{30}&=2265,\qquad S_{31}=2418.\end{aligned}

For positive n, the factor 5n+151 is positive. Equality at n=30 is insufficient.

How did this part go?

Solution notes

The printed answers are a=3,d=5 and least n=31; the algebra supplements the answer key.