As    a_n = 3 + 2n,
so    a₁ = 3 + 2 = 5
 a₂ = 3 + 2 x 2 = 7
a₃ = 3 + 2 x 3 = 9
List of numbers becomes 5, 7, 9, 11, . . .
Here,   7 – 5 = 9 – 7 = 11 – 9 = 2 and so on.
So, it forms an AP with common difference d = 2.
To find S₂₄, we have n = 24, a = 5, d = 2.
Therefore, \(\mathrm{S}_{24}=\frac{24}{2}[2 \times 5+(24-1) \times 2]=12[10+46]=672\)
So, sum of first 24 terms of the list of numbers is 672.

LHS = \({ (\sin { A } +cosecA) }^{ 2 }+{ (\cos { A } +\sec { A } ) }^{ 2 }\)
\(=\sin ^{ 2 }{ A } +{ cosec }^{ 2 }A+2\sin { A } cosecA+\cos ^{ 2 }{ A } +\sec ^{ 2 }{ A } +2\cos { A } \sec { A } \)
\(\left[ \because { (a+b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab \right] \)
\(=(\sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } )+(1+\cot ^{ 2 }{ A } )+2\sin { A } \frac { 1 }{ \sin { A } } +(1+\tan ^{ 2 }{ A } )+2\cos { A } \frac { 1 }{ \cos { A } } \)\(\left[ \because { cosec }^{ 2 }A=1+\cot ^{ 2 }{ A } ,\sec ^{ 2 }{ A } =1+\tan ^{ 2 }{ A } ,cosecA=\frac { 1 }{ \sin { A } } and\sec { A } =\frac { 1 }{ \cos { A } } \right] \)\(=1+1+\cot ^{ 2 }{ A } +2+1+\tan ^{ 2 }{ A } +2\quad \left[ \because \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } =1 \right] \)
\(=7+\tan ^{ 2 }{ A } +\cot ^{ 2 }{ A } \)
= RHS
Hence proved.

\(\sec { \theta } =\frac { 4 }{ \sqrt { 17 } } =\frac { H }{ B } \)
Using Pythagoras theorem,
P²=H²-B²=16-7=9
\(\Rightarrow\)P=3
Now, \(\tan { \theta } =\frac { P }{ B } =\frac { 3 }{ \sqrt { 7 } } \) ,
\(cosec\theta =\frac { H }{ P } =\frac { 4 }{ 3 } ,\cos { \theta } =\frac { B }{ H } =\frac { \sqrt { 7 } }{ 4 } ,\cot { \theta } =\frac { \sqrt { 7 } }{ 3 } \)
Now, put these values in LHS of given expression.

As, \(\sqrt { 3 } \sin { \theta } =\cos { \theta } \Rightarrow \tan { \theta } =\frac { 1 }{ \sqrt { 3 } } \)
\(\therefore \frac { \sin { \theta } \tan { \theta } (1+\cot { \theta } ) }{ \sin { \theta } +\cos { \theta } } =\frac { \tan { \theta } \tan { \theta } (1+\cot { \theta } ) }{ \tan { \theta } +1 } \)
\(\frac { 1+\sqrt { 3 } }{ 4 } \)

10

Remaining cards of spade=13-3=10
So, favourable outcomes=10,
i.e. n(E₄)=10
∴ P(getting a spade)​​​​​​​\(=\frac { 10 }{ 49 } \)

Let Ist term of AP be a and common difference be d.
Now, three middle terms of this AP are a₁₀, a₁₁ and a₁₂
A.T.Q.,  a₁₀ + a₁₁ + a₁₂ = 129
\(\Rightarrow\)  ( a + 9d ) + ( a + 10d ) + ( a + 11d ) = 129
\(\Rightarrow\)    3a + 30d = 129
\(\Rightarrow\)     a + 10d = 43  \(\Rightarrow\)  a = 43 - 10d       ..(i)
Also,  last three terms are a₁₉, a₂₀ and a₂₁
\(\therefore\)   a₁₉ + a₂₀ + a₂₁ = 237
\(\Rightarrow \)  ( a + 18d ) + ( a + 19d ) + ( a + 20d )= 237
\(\Rightarrow\)    3a + 57d = 237   \(\Rightarrow\)  a + 19d = 79
\(\Rightarrow\)  43 - 10d + 19d = 79               [ Using eq.(i) ]
\(\Rightarrow\)  9d = 36  \(\Rightarrow\)  d = 4
When d = 4, equation (i) becomes
a = 43 - 10 x 4 = 3
\(\therefore\)   AP is 3, 7, 11, 15, ...