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- Question 4 & 5 Exercise 5.1
- $ terms. ====Solution==== The general term of the sequence is: \begin{align}& T_j=\dfrac{j}{2}[2(2)+3(j-1)]\
- Question 1 Exercise 5.3
- cdots \\ & a_n-a_{n-1}=(n-1)th \quad\text{term of sequence}\quad 9,15,21,...\end{align} Which is in A.P. col
- Question 2 Exercise 5.3
- \\ & a_n-a_{n-1}=(\mathrm{n}-1)\text{ term of the sequence} 10,16,22, \ldots\end{align} which is a A.P. Addi
- Question 3 Exercise 5.3
- -a_{n \quad 1}=(\mathrm{n}-1) \text { term of the sequence } \end{align} $6,10,8, \ldots$ which is a A.P. Ad
- Question 4 Exercise 5.3
- \ & a_n-a_{n-1}=(\mathrm{n}-1) \text { term ofthe sequence }\end{align} $6,10,18, \ldots$ which is a G.P. Ad
- Question 5 Exercise 5.3
- _n-a_{n-1}=(\mathrm{n}-1)\quad \text{ term of the sequence}\quad 6,12,24, \ldots\end{align} which is a G.P.
- Question 6 Exercise 5.3
- \ & a_n-a_{n-1}=(\mathrm{n}-1) \text { term ofthe sequence } 4,20,100, \ldots \end{align} which is a G.P wit
- Question 2 & 3 Review Exercise
- term is the product of corresponding terms of the sequences $1,2.3, \ldots, 3,4,5 \ldots$ and $5,6,7, \ldots
- Question 4 Review Exercise
- erm is the product of the successive terms of the sequence $1,4,7, \ldots$ Thus the general term of the ser