Here, a₁, a₂, a₃, ....., a_n be the given A.P. with a1 as first term and d as the common difference
\(\therefore\)    a₂ = a₁ + d
a₃ = a₁ + 2d
and  a_n = a₁ + ( n - 1 )d
Now,    L.H.S = \(\frac{1}{{a}_{1}{a}_{2}}+\frac{1}{{a}_{2}{a}_{3}}+...+\frac{1}{{a}_{n-1}{a}_{n}}\)
\(=\frac{1}{{a}_{1}({a}_{1}+d)}+\frac{1}{({a}_{1}+d)({a}_{1}+2d)}+...+\frac{1}{[{a}_{1}+(n-2)d][{a}_{1}+(n-1)d]}\)

\(\frac{1}{d}\left[ \frac{1}{{a}_{1}}-\frac{1}{{a}_{1}+d} \right]+\frac{1}{d}\left[ \frac{1}{{a}_{1}+d} -\frac{1}{{a}_{1}+2d}\right]+....+\frac{1}{d}\left[ \frac{1}{{a}_{1}+(n-2)s}-\frac{1}{{a}_{1}+(n-1)d} \right]\)
\(-=\frac{1}{d}\left[ \frac{1}{{a}_{1}}-\frac{1}{{a}_{1}+d}+\frac{1}{{a}_{1}+d}-\frac{1}{{a}_{1}+2d} +....+\frac{1}{{a}_{1}+(n-2)d} -\frac{1}{{a}_{1}+(n-1)d}\right]\)
\(=\frac{1}{d}\left[ \frac{1}{{a}_{1}}-\frac{1}{{a}_{1}+(n-1)d} \right]\)
\(=\frac{1}{d}\left[ \frac{{a}_{1}+(n-1)d-{a}_{1}}{{a}_{1}({a}_{1}+(n-1)d)} \right]\)
\(=\left[ \frac{(n-1)d}{{a}_{1}{a}_{n}} \right]\)
\(=\frac{n-1}{{a}_{1}{a}_{n}}\) = R.H.S.

66.67m

Steps of Construction : 
1. Draw a line segment PQ = 4 cm. 
2. At Q, construct an ㄥPQY = 60^o.
3. Through P, draw an arc of radius 3.6 cm, intersecting QY in S.
4. With S as centre and radius equal to SQ, draw an arc to intersect QY in R.
5. Join PR to get ΔPQR.
6. At P, construct an acute angle ㄥQPX ( < 90^o).
7. Mark four points on PX such that PP₁ = P₁P₂ = P₂P₃ = P₃P₄.
8. Join P₃Q.
9. Through P₄, draw P₄Q' || P₃Q, intersecting PQ in Q', when produced.
10. Through Q', draw Q'R' || QR intersecting PR in PR', when produced.
Thus, ΔPO'R' is the required triangle.

Steps of Construction : 
1. Draw a line segment AB = 13.5 cm.
2. Through A, construct an acute angle ㄥBAX ( < 90^o).
3. Mark nine points (2 + 3 + 4 = 9) at equal distances on AX Such that AA₁= A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ =  A₅A₆ = A₆A₇ = A₇A₈ = A₈A₉.
4. Join A₉B.
5. Through A₂ and A₅ draw A₅R || A₉B and A₂Q || A₉B, intersecting AB in Q and R.
6. With Q as centre draw an arc of radius AQ.
7. With R as centre draw another arc of radius RB, intersecting previous arc in P.
8. Join PQ and PR

Thus, ΔPQR is the required triangle.

9.72 cm²

30.38 cm²

49.28 cm²

Suppose R and r be the radii of bigger and smaller circles, respectively
⇒ AB - AC = CB
⇒ 2R-2r=9
⇒ R-r=\(9\over 2\)= 4.5 cm ...(i)  

Join AD and CD
ΔAOD ~ ΔDOC
\(⇒\ {OD\over OA}={OC\over OD}\)
⇒ OD² = OA x OC
⇒ (R-5)²=R x (R-9)
⇒R²+25-10R=R²-9R
⇒ R=25cm
From (i), we have R-r = 4.5
=25 - 4.5 = 20.5cm
Now, area of the shaded portion = πR²-π​​​​​​​r²
=π​​​​​​​(R²-r²)
=π​​​​​​​(R+r)(R-r)
=3.14 x (25+20.5)(25-20.5)
=3.14 x 45.5 x 4.​​​​​​​5
= 642.915 cm²
Hence, the required area of the shaded portion is 642.915cm²

Given, height of the cone=8 cm and radius of the cone = 5 cm

Volume of cone=\(\frac{1}{3}\)πr²h
=\(\frac{1}{3}\)π(5)²8 cm³=\(\frac{200}{3}\)π cm³
Radius of one spherical lead shot=0.5 cm
Volume of one spherical lead shot=\(\frac{4}{3}\)πr ³=\(\frac{4}{3}\)π(0.5)³ cm³
=\(\frac{4\times{0.125}}{3}\)π cm³=\(\frac{0.5}{3}\)π cm³
When spherical lead are dropped in the vessel, one fourth of water flows out
Let number of lead shots be n
Volume of n spherical shots = \(\frac{1}{4}\) volume of conical vessel
n\(\left(\frac{0.5}{3}\pi\right)\)=\(\frac{1}{4}\left(\frac{200}{3}\pi\right)\)⇒n(0.5)=50⇒ n=\(\frac{50\times{10}}{5}\)=100 

Let G₁, G₂, G₃ and B₁, B₂ be three girls and two boys respectively.
Since two children are selected at random, therefore the following are the possible groups:
B₁B₂, B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₃, G₁G₂, G₁G₃, G₂G₃
Total number of cases = 10
Now, at least one boy is selected in the following ways:
B₁B₂, B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₃
Total number of favourable cases = 7  
\(\therefore\) Required probability = \(​​\frac {7} {10}\)