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- Question 5 and 6, Exercise 4.2
- = -73 \quad \cdots (2) \end{align*} Now, subtract equation (1) from equation (2): \begin{align*} \begin{array}{ccc} a_1& + 27d &= -73\\ \mathop{}\limits_{-}a_1 &\ma... = 43 \quad \cdots (2) \end{align*} Now, subtract equation (1) from equation (2): \begin{align*} \begin{array}{ccc} a_1 & + 10d &= 43\\ \mathop{}\limits_{-}a_1 &\math
- Question 9 and 10, Exercise 4.8
- 2) \end{align*} Now, put $k+1=0 \implies k=-1$ in equation (2): \begin{align*} 1 &= (-1+2)A + 0 \\ \implies ... . \end{align*} Next, put $k+2=0 \implies k=-2$ in equation (2): \begin{align*} 1 &= 0 + (-2+1)B \\ \implies ... . \end{align*} Using the values of $A$ and $B$ in equation (1), we get \begin{align*} \frac{1}{(k+1)(k+2)} &
- Question 13, 14 and 15, Exercise 4.8
- Now, put $2k+3 = 0 \implies k = -\frac{3}{2}$ in equation (2): \begin{align*} 1 &= (2 \times \left(-\frac{3... Next, put $2k+9 = 0 \implies k = -\frac{9}{2}$ in equation (2): \begin{align*} 1 &= 0 + (2 \times \left(-\fr... . \end{align*} Using the values of $A$ and $B$ in equation (1), we get \begin{align*} \frac{1}{(2k+3)(2k+9)}
- Question 3 and 4, Exercise 4.2
- = -1 \quad \cdots (2) \end{align*} Now, subtract equation (1) from equation (2): \begin{align*} \begin{array}{ccc} a_1 & + 8d &= -1\\ \mathop{}\limits_{-}a_1 &\matho
- Question 11 and 12, Exercise 4.8
- . \end{align*} Using the values of $A$ and $B$ in equation (1), we get \begin{align*} \frac{1}{k(k+2)} &= \f