(a.i) If [tex]a_1,a_2,a_3,\ldots[/tex] are in arithmetic progression, then there is a constant [tex]d[/tex] such that
[tex]a_{n+1} - a_n = d[/tex]
for all [tex]n\ge1[/tex]. In other words, the difference [tex]d[/tex] between any two consecutive terms in the sequence is always the same.
[tex]a_2-a_1 = a_3-a_2 = a_4-a_3 = \cdots = d[/tex]
Now, we can expand the target expression into partial fractions.
[tex]\dfrac1{a_na_{n+1}} = \dfrac{\alpha}{a_n} + \dfrac{\beta}{a_{n+1}}[/tex]
Combining the fractions on the right and using the recursive equation above, we have
[tex]\dfrac1{a_na_{n+1}} = \dfrac{\alpha (a_n + d) + \beta a_n}{a_n(a_n+d)} = \dfrac{(\alpha+\beta) a_n + \alpha d}{a_n a_{n+d}} \\\\ \implies \begin{cases}\alpha + \beta = 0 \\ \alpha d = 1 \end{cases} \implies \alpha = \dfrac1d, \beta = -\dfrac1d[/tex]
and hence
[tex]\dfrac1{a_n a_{n+1}} = \dfrac1{da_n} - \dfrac1{da_{n+1}}[/tex]
as required.
(a.ii) Using the previous result, the [tex]n[/tex]-th term [tex](n\ge1)[/tex] in the sum on the left is
[tex]\dfrac1{a_n a_{n+1}} = \dfrac1d \left(\dfrac1{a_n} - \dfrac1{a_{n+1}}\right)[/tex]
Expand each term in this way to reveal a telescoping sum:
[tex]\dfrac1{a_1a_2} + \dfrac1{a_2a_3} + \dfrac1{a_3a_4} + \cdots + \dfrac1{a_{99}a_{100}} \\\\ ~~~~~~~~ = \dfrac1d \left(\left(\dfrac1{a_1} - \dfrac1{a_2}\right) + \left(\dfrac1{a_2} - \dfrac1{a_3}\right) + \left(\dfrac1{a_3} - \dfrac1{a_4}\right) + \cdots + \left(\dfrac1{a_{99}} - \dfrac1{a_{100}}\right)\right) \\\\ ~~~~~~~~ = \dfrac1d \left(\dfrac1{a_1} - \dfrac1{a_{100}}\right) = \dfrac{a_{100} - a_1}{d a_1 a_{100}}[/tex]
By substitution, we can show
[tex]a_n = a_{n-1} + d = a_{n-2} + 2d = \cdots = a_1 + (n-1)d \\\\ \implies a_{100} = a_1 + 99d[/tex]
so that the last expression reduces to
[tex]\dfrac{(a_1 + 99d) - a_1}{d a_1 a_{100}} = \dfrac{99d}{d a_1 a_{100}} = \dfrac{99}{a_1 a_{100}}[/tex]
as required. More generally, it's easy to see that
[tex]\dfrac1{a_1a_2} + \dfrac1{a_2a_3} + \dfrac1{a_3a_4} + \cdots + \dfrac1{a_na_{n+1}} = \dfrac{n}{a_1a_{n+1}}[/tex]
(b) I assume you mean the equation
[tex]\dfrac1{3\times7} + \dfrac1{7\times11} + \dfrac1{11\times15} + \cdots + \dfrac1{k(k+4)} = \dfrac2{25}[/tex]
Note that the distinct factors of each denominator on the left form an arithmetic sequence.
[tex]a_1 = 3[/tex]
[tex]a_2 = 3 + 4 = 7[/tex]
[tex]a_3 = 7 + 4 = 11[/tex]
and so on, with [tex]n[/tex]-th term
[tex]a_n = 3 + (n-1)\times4 = 4n - 1[/tex]
Let [tex]a_n=k[/tex]. Using the previous general result, the left side reduces to
[tex]\dfrac1{3\times7} + \dfrac1{7\times11} + \dfrac1{11\times15} + \cdots + \dfrac1{a_na_{n+1}} = \dfrac n{3a_{n+1}} \\\\ \implies \dfrac{\frac{k+1}4}{3(k+4)} = \dfrac2{25}[/tex]
Solve for [tex]k[/tex].
[tex]\dfrac{k+1}{12k+48} = \dfrac2{25} \implies 25(k+1) = 2(12k+48) \\\\ \implies 25k + 25 = 24k + 96 \implies \boxed{k=71}[/tex]
