Question

This is an Ap question anyone solve plz.Give explanation as well

Answer 1

$\underline{\underline{\Large{\mathfrak{Solution : }}}}$



$\textsf{Here each term is the product of two different}\\ \textsf{arithmetic progression.}$




$\underline{\textsf{For 1st A.P,}} \\ \\ \mathsf{\implies First \: term( a_1) \: = \: 1 } \\ \\ \mathsf{\implies Common \: difference (d_1) \: = \: 3 \: - \: 1 \: = \: 2 } \\ \\ \\ \underline{\textsf{For 2nd A.P,}} \\ \\ \mathsf{\implies First \: term (a_2) \: = \: 3 } \\ \\ \mathsf{\implies Common \: difference ( d_2) \: = \: 5 \: - \: 3 \: = \: 2 }$



$\underline{\textsf{Now,}} \\ \\ \mathsf{\implies T_n \: = \: (T_n)_1 \: \times \: (T_n)_2}$



$\mathsf{\implies T_n \: = \: \{ a_1 \: + \: ( n \: - \: 1)d_1 \: \}\{ a_2 \: + \: ( n \: - \: 1)d_2\}} \\ \\ \mathsf{\implies T_n \: = \: \{ 1 \: + \: ( n \: - \: 1)2 \}\{3 \: + \: ( n \: - \: 1)2\}}$



$\mathsf{\implies T_n \: = \: \{ 1 \: + \: 2n \: - \: 2 \}\{3 \: + \: 2n \: - \: 2\}} \\ \\ \mathsf{\implies T_n \: = \: \{ 2n \: - \: 1\}\{2n \: + \: 1 \}} \\ \\ \\ \mathsf{\: \: \therefore \: T_n \: = \: 4n^2 \: - \: 1 }$



$\underline{\textsf{Now,}} \\ \\ \mathsf{\implies S_n \: = \: \sum_{n \: = \: 1}^{n} T_n}$



$\mathsf{\implies S_n \: = \: \sum_{n \: = \: 1}^{n} ( 4n^2 \: - \: 1 ) } \\ \\ \mathsf{\implies S_n \: = \: \sum_{n\: = \: 1}^{n} 4n^2 \: - \: n } \\ \\ \mathsf{\implies S_n \: = \: 4\sum_{n \: = \: 1}^{n}n^2 \: - \: n }$



$\mathsf{\implies S_n \: = \: 4 \: \times \: \dfrac{n(n \: + \: 1)(2n \: + \: 1)}{6} \: - \: n}$



$\mathsf{\implies S_{100} \: = \: \dfrac{4 \: \times \: 100 \: ( 100 \: + \: 1)( 2 \: \cdot \: 100 \: + \: 1 )}{6} \: - \: 100 }$


$\mathsf{\implies S_{100} \: = \: \dfrac{ 4 \: \cdot \: 100 \: \cdot \: 101 \: \cdot \: 201}{6} \: - \: 100}$



$\mathsf{\implies S_{100} \: = \: 2 \: \cdot \: 100 \: \cdot \: 101 \: \cdot \: 67 \: - \: 100 } \\ \\ \mathsf{\implies S_{100} \: = 1,353,400 \: - \: 100}$


$\mathsf{\: \therefore \: \: S_{100} \: = \: 1,353,300}$

Answer 2

$\underline{\underline{\Large{\mathfrak{Solution : }}}} \\\\\\</p><p></p><p> </p><p> </p><p> </p><p> </p><p></p><p></p><p></p><p></p><p>\begin{lgathered}\textsf{Here each term is the product of two different}\\ \textsf{arithmetic progression.}\end{lgathered} </p><p>\\\\\\</p><p> </p><p> </p><p></p><p></p><p></p><p></p><p>\begin{lgathered}\underline{\textsf{For 1st A.P,}} \\ \\\mathsf{\implies First \: term( a_1) \: = \: 1 } \\ \\\mathsf{\implies Common \: difference (d_1) \: = \: 3 \: - \: 1 \: = \: 2 } \\ \\ \\ \underline{\textsf{For 2nd A.P,}} \\ \\ \mathsf{\implies First \: term (a_2) \: = \: 3 } \\ \\ \mathsf{\implies Common \: difference ( d_2) \: = \: 5 \: - \: 3 \: = \: 2 }\end{lgathered} \\\\\\</p><p></p><p> </p><p> </p><p></p><p></p><p></p><p></p><p></p><p>\begin{lgathered}\underline{\textsf{Now,}} \\ \\ \mathsf{\implies T_n \: = \: (T_n)_1 \: \times \: (T_n)_2}\end{lgathered} </p><p>\\\\\\</p><p> </p><p> </p><p></p><p></p><p>\begin{lgathered}\mathsf{\implies T_n \: = \: \{ a_1 \: + \: ( n \: - \: 1)d_1 \: \}\{ a_2 \: + \: ( n \: - \: 1)d_2\}} \\ \\ \mathsf{\implies T_n \: = \: \{ 1 \: + \: ( n \: - \: 1)2 \}\{3 \: + \: ( n \: - \: 1)2\}}\end{lgathered} \\\\\\</p><p></p><p> </p><p> </p><p></p><p></p><p></p><p>\begin{lgathered}\mathsf{\implies T_n \: = \: \{ 1 \: + \: 2n \: - \: 2 \}\{3 \: + \: 2n \: - \: 2\}} \\ \\ \mathsf{\implies T_n \: = \: \{ 2n \: - \: 1\}\{2n \: + \: 1 \}} \\ \\ \\ \mathsf{\: \: \therefore \: T_n \: = \: 4n^2 \: - \: 1 }\end{lgathered} \\\\\\</p><p></p><p> </p><p> </p><p></p><p></p><p>\begin{lgathered}\underline{\textsf{Now,}} \\ \\ \mathsf{\implies S_n \: = \: \sum_{n \: = \: 1}^{n} T_n}\end{lgathered} \\\\\\</p><p></p><p> </p><p> </p><p></p><p></p><p></p><p>\begin{lgathered}\mathsf{\implies S_n \: = \: \sum_{n \: = \: 1}^{n} ( 4n^2 \: - \: 1 ) } \\ \\ \mathsf{\implies S_n \: = \: \sum_{n\: = \: 1}^{n} 4n^2 \: - \: n } \\ \\ \mathsf{\implies S_n \: = \: 4\sum_{n \: = \: 1}^{n}n^2 \: - \: n }\end{lgathered} \\\\\\</p><p></p><p></p><p></p><p>\mathsf{\implies S_n \: = \: 4 \: \times \: \dfrac{n(n \: + \: 1)(2n \: + \: 1)}{6} \: - \: n}\\\\\\</p><p></p><p></p><p></p><p>\mathsf{\implies S_{100} \: = \: \dfrac{4 \: \times \: 100 \: ( 100 \: + \: 1)( 2 \: \cdot \: 100 \: + \: 1 )}{6} \: - \: 100 }\\\\\\</p><p></p><p></p><p>\mathsf{\implies S_{100} \: = \: \dfrac{ 4 \: \cdot \: 100 \: \cdot \: 101 \: \cdot \: 201}{6} \: - \: 100}\\\\\\</p><p></p><p>\begin{lgathered}\mathsf{\implies S_{100} \: = \: 2 \: \cdot \: 100 \: \cdot \: 101 \: \cdot \: 67 \: - \: 100 } \\ \\ \mathsf{\implies S_{100} \: = 1,353,400 \: - \: 100}\end{lgathered} \\\\\\</p><p></p><p> </p><p> </p><p></p><p></p><p>\mathsf{\: \therefore \: \: S_{100} \: = \: 1,353,300}$