Question

Solve the ques 6. ( i ) and ( ii ) in the above attachment

Answer 1

$\underline{\underline{\large{\mathfrak{Solution : }}}} \\ \\ \\ \underline{\textsf{Question no. 6(i) : }}$


$\underline{\textsf{Given : }} \\ \\ \mathsf{\implies S_{n} \: = \: \Bigg( \dfrac{5n^{2}}{2} \: + \: \dfrac{3n}{2} \Bigg)} \\ \\ \underline{\textsf{To Find : }} \\ \\ \mathsf{\implies T_{n} \: = \: ? } \\ \\ \mathsf{\implies T_{20} \: = \: ?}$




$\textsf{We know that : } \\ \\ \mathsf{\implies T_{n} \: = \: S_{n} \: - \: S_{(n \: - \: 1)}}$



$\mathsf{\implies T_{n} \: = \: \Bigg( \dfrac{5n^{2}}{2} \: + \: \dfrac{3n}{2} \Bigg) \: - \: \Bigg\{ \dfrac{5 {(n \: - \: 1)}^{2} }{2}\: + \: \dfrac{3(n \: - \: 1)}{2} \Bigg\}}$





$\mathsf{\implies T_{n} \: = \: \Bigg( \dfrac{5n^{2} \: + \: 3n}{2} \Bigg) \: - \: (n \: - \: 1)\Bigg\{ \dfrac{5(n \: - \: 1)}{2} \: + \: \dfrac{3}{2} \Bigg\}}$




$\mathsf{\implies T_{n} \: = \: \Bigg( \dfrac{5n^{2} \: + \: 3n}{2} \Bigg) \: - \: (n \: - \: 1) \Bigg\{ \dfrac{5(n \: - \: 1) \: + \: 3}{2} \Bigg\} }$



$\mathsf{\implies T_{n} \: = \: \Bigg( \dfrac{5n^{2} \: + \: 3n}{2} \Bigg) \: - \: (n \: - \: 1) \Bigg(\dfrac{5n \: - \: 5\: + \: 3}{2} \Bigg) }$



$\mathsf{\implies T_{n} \: = \: \Bigg( \dfrac{5n^{2} \: + \: 3n}{2} \Bigg) \: - \: (n \: - \: 1) \Bigg(\dfrac{5n \: - \: 2}{2} \Bigg)}$




$\mathsf{\implies T_{n} \: = \: \Bigg\{ \dfrac{(5n^{2} \: + \: 3n) \: - \: (n \: - \: 1)(5n \: - \: 2)}{2}\Bigg\}}$



$\mathsf{\implies T_{n} \: = \: \Bigg\{ \dfrac{(5n^{2} \: + \: 3n) \: - \: (5 {n}^{2} \: - \: 2n \: - \: 5n \: + \: 2 )}{2}\Bigg\}}$



$\mathsf{\implies T_{n} \: = \: \Bigg\{ \dfrac{ \cancel{5n^{2}} \: + \: 3n \: - \: \cancel{5 {n}^{2}} \: + \: 2n \: + \: 5n \: - \: 2 }{2}\Bigg\}}$



$\mathsf{\implies T_{n} \: = \: \Bigg\{ \dfrac{ 10n \: - \: 2}{2} \Bigg\}} \\ \\ \\ \mathsf{\implies T_{n} \: = \: \Bigg\{ \dfrac{ \cancel{2}(5n \: - \: 1)}{\cancel{2}} \Bigg\}} \\ \\ \\ \mathsf{\therefore \quad \: T_{n} \: = \: 5n \: - \: 1}$



$\underline{\mathsf{To \: Find \longrightarrow 20th \: term : }} \\ \\ \mathsf{\implies T_{n} \: = \: 5n \: - \: 1} \\ \\ \mathsf{\implies T_{20} \: = \: 5 \: \times \: 20 \: - \: 1} \\ \\ \mathsf{\implies T_{20} \: = \: 100 \: - \: 1} \\ \\ \mathsf{\therefore \quad \: T_{20} \: = \: 99}$





$\underline{\textsf{Question no. 6(ii) : }}$




$\underline{\textsf{Given : }} \\ \\ \mathsf{\implies S_{n} \: = \: \Bigg( \dfrac{3n^{2}}{2} \: + \: \dfrac{5n}{2} \Bigg)} \\ \\ \underline{\textsf{To Find : }} \\ \\ \mathsf{\implies T_{n} \: = \: ? } \\ \\ \mathsf{\implies T_{25} \: = \: ?}$




$\textsf{We know that : } \\ \\ \mathsf{\implies T_{n} \: = \: S_{n} \: - \: S_{(n \: - \: 1)}}$




$\mathsf{\implies T_{n} \: = \: \Bigg( \dfrac{3n^{2}}{2} \: + \: \dfrac{5n}{2} \Bigg) \: - \: \Bigg\{ \dfrac{3 {(n \: - \: 1)}^{2} }{2}\: + \: \dfrac{5(n \: - \: 1)}{2} \Bigg\}}$




$\mathsf{\implies T_{n} \: = \: \Bigg( \dfrac{3n^{2} \: + \: 5n}{2} \Bigg) \: - \: (n \: - \: 1)\Bigg\{ \dfrac{3(n \: - \: 1)}{2} \: + \: \dfrac{5}{2} \Bigg\}}$



$\mathsf{\implies T_{n} \: = \: \Bigg( \dfrac{3n^{2} \: + \: 5n}{2} \Bigg) \: - \: (n \: - \: 1) \Bigg\{ \dfrac{3(n \: - \: 1) \: + \: 5}{2} \Bigg\} }$



$\mathsf{\implies T_{n} \: = \: \Bigg( \dfrac{3n^{2} \: + \: 5n}{2} \Bigg) \: - \: (n \: - \: 1) \Bigg(\dfrac{3n \: - \: 3\: + \: 5}{2} \Bigg) }$



$\mathsf{\implies T_{n} \: = \: \Bigg( \dfrac{3n^{2} \: + \: 5n}{2} \Bigg) \: - \: (n \: - \: 1) \Bigg(\dfrac{3n \: + \: 2}{2} \Bigg)}$



$\mathsf{\implies T_{n} \: = \: \Bigg\{ \dfrac{(3n^{2} \: + \: 5n) \: - \: (n \: - \: 1)(3n \: + \: 2)}{2}\Bigg\}}$



$\mathsf{\implies T_{n} \: = \: \Bigg\{ \dfrac{(3n^{2} \: + \: 5n) \: - \: (3 {n}^{2} \: + \: 2n \: - \: 3n \: - \: 2 )}{2}\Bigg\}}$



$\mathsf{\implies T_{n} \: = \: \Bigg\{ \dfrac{ \cancel{3n^{2}} \: + \: 5n \: - \: \cancel{3 {n}^{2}} \: - \: 2n \: + \: 3n \: + \: 2 }{2}\Bigg\}}$



$\mathsf{\implies T_{n} \: = \: \Bigg\{ \dfrac{ 6n \: + \: 2}{2} \Bigg\}} \\ \\ \\ \mathsf{\implies T_{n} \: = \: \Bigg\{ \dfrac{ \cancel{2}(3n \: + \: 1)}{\cancel{2}} \Bigg\}} \\ \\ \\ \mathsf{\therefore \quad \: T_{n} \: = \: 3n \: + \: 1}$



$\underline{\mathsf{To \: Find \longrightarrow 25th \: term : }} \\ \\ \mathsf{\implies T_{n} \: = \: 3n \: + \: 1} \\ \\ \mathsf{\implies T_{25} \: = \: 3 \: \times \: 25 \: + \: 1} \\ \\ \mathsf{\implies T_{25} \: = \: 75\: + \: 1} \\ \\ \mathsf{\therefore \quad \: T_{25} \: = \: 76}$

Answer 2

Hey there !!


↪ In question no. (6) → [i] .

➡ Given :-

→ S$\tiny n$ = $\frac{5 {n}^{2} }{2} + \frac{3n}{2}$


➡ To find :-

→ nth term [ a$\tiny n$ ] .

→ 20th term [ a$\tiny 20$ ] .


➡ Solution :-


▶See the attachment 1 and 2.



↪ In question no. (6) → [ii] .

➡ Given :-

→ S$\tiny n$ = $\frac{3 {n}^{2} }{2} + \frac{5n}{2} .$

➡ To find :-

→ nth term [ a$\tiny n$ ] .

→ 25th term [ a$\tiny 25$ ] .


➡ Solution :-

▶ See the attachment 2 and 3.


✔✔ Hence, it is solved ✅✅.

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THANKS

#BeBrainly.