Live practice

Formula or theorem used: Mark the corresponding sides and angles before applying the Basic Proportionality Theorem or an AAA, SAS, or SSS similarity test.

Working: State the theorem or similarity criterion, write equal ratios in corresponding order, substitute the given lengths, and end with the required equality or length.

After substitution and simplification:

\(42\,\mathrm{m}\)

Here \(a=9\) and \(d=8\). Using (S_n=\frac n2[2a+\(n-1\)d]),

The positive solution is \(n=12\).

Since \(\triangle ABC\sim\triangle PQR\),

Because \(AD\) and \(PM\) are medians, \(BD=B\frac{C}{2}\) and \(QM=Q\frac{R}{2}\). Therefore,

Also, \(\angle ABD=\angle PQM\), because \(D\) lies on \(BC\), \(M\) lies on \(QR\), and the corresponding full-triangle angles are equal.

Thus \(\triangle ABD\sim\triangle PQM\) by SAS similarity. Corresponding sides give

Using (S_n=\frac n2(a+l)),

Now (l=a+\(n-1\)d):

Therefore,

From \(a_{22}=a+21d\),

Hence,

Subtracting gives \(d=4\), and then \(a=10\). Thus,

Using (S_n=\frac n2[2a+\(n-1\)d]),

Subtracting \(1\) from \(2\), \(5d=10\), so \(d=2\), and \(1\) gives \(a=1\). Therefore,

\(i\) (a_{n+1}-a_n=[3+4\(n+1\)]-\(3+4n\)=4), a constant. Thus it is an AP with \(a_1=7\), \(d=4\).

\(ii\) (a_{n+1}-a_n=[9-5\(n+1\)]-\(9-5n\)=-5), a constant. Thus \(a_1=4\), \(d=-5\).

(S_1=4\(1\)-1^2=3) and (S_2=4\(2\)-2^2=4).

For \(n\ge2\), \(a_n=S_n-S_{n-1}\). Hence,

Therefore \(a_2=1\), \(a_3=-1\), and \(a_{10}=-15\).