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- Question 1 Exercise 5.1
- ldots$ whose $n^{\text {th }}$ term is $2 n-1$. Therefore $n^{t h}$-term of - the given series is: $$T... }$ term is: $a_j=1+4(j-1)=4 j-3$. \begin{align}& \therefore T_j-(4 j-3)^3 \\ & =64 j^3-144 j^2+108 j-27\e
- Question 1 Exercise 5.3
- of two successive terms of the A.P 2,5,8,11,... Therefore, the general term of the series is: $$u_n=\df... o successive terms of the A.P $4.13,22, \ldots$. Therefore, the general term of the series is: $u_n=\dfr
- Question 2 & 3 Review Exercise
- \dfrac{2 n+4}{3} \\ & =\dfrac{n(n+1)(n+2)}{3} \\ \therefore S_n &=\sum_{r=1}^n a_r=\dfrac{n(n+1)(n+2)}{3}... h}$ term $n, \quad n+2$ and $n+4$ respectively. Therefore, $n^{t h}$ term of the series is $a_n=n(n+2)(
- Question 7 Review Exercise
- ts,(2 n-1)$ and $2^2, 3^2, 4^2, \ldots,(n+1)^2$. Therefore, the general lerm of the series is: \begin{al... ldots,(2 n-1)$ and $1^2, 2^2, 3^2, \ldots, n^2$. Therefore, the $n^{\text {th }}$ term of the given seri
- Question 2 & 3 Exercise 5.1
- are $n(n+1)$ and the given series have 99 terms. Therefore, the $n^{\text {th }}$ term of the given seri
- Question 6 Exercise 5.1
- {t h}$ terms are $j, j+1$ and $j+2$ respectively, therefore the $n^{t h}$ term of the given series is: \b
- Question 4 & 5 Exercise 5.2
- given $S_{\infty}=\dfrac{44}{9}$ \begin{align} & \therefore 3+\dfrac{2 r}{1-r}=\dfrac{44}{9} \\ & \Righta
- Question 9 Review Exercise
- ch is a geometric series with common ratio $r=3$. Therefore, \begin{align} a_n-a_1&=\dfrac{3(3^{n-1}-1)}{