Since 2k-7, k + 5 and 3k + 2 are three consective terms of an A.P.
k + 5 - (2k - 7) = 3k + 2 - (k + 5)
k + 5 - 2k + 7 = 3k + 2- k- 5
- k + 12 = 2k - 3
3k = 15
k = 5.

Since the sum of the first n terms of an A.P. is given as 3n² + 5n
\(\therefore\)  S_n = 3n² + 5n                ....(i)
\(\therefore\)   S_n-1 = 3( n - 1)2 + 5 ( n - 1 ) 
= 3( n² + 1 - 2n ) + 5( n - 1 )
= 3n² + 3 - 6n + 5n - 5 
= 3n² - n - 2                  ...(ii)
Now,  an = S_n - S_n-1
= 3n² + 5n - 3n² + n + 2
an =  6n + 2
Put n = 1 and n = 2, we have
a₁ = 8 and a₂ = 14
Hence, the common difference is 6.

Let a and d be the first term and common difference of given A.P. respectively.
\(\therefore\)   Sm = \(\frac{m}{2}[2a+(m-1)d]\)
Sn = \(\frac{n}{2}[2a+(m-1)d]\)
According to the statement of the question
S_m : S_n = m² : n²
i.e, \(\frac{\frac{m}{2}]2a+(m-1)d]}{\frac{m}{2}]2a+(n-1)d]}=\frac{{m}^{2}}{{n}^{2}}\)
\(\Rightarrow\)       \(\frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}\)               ...(i)
Now,       \(\frac{{a}_{m}}{{a}_{n}}=\frac{a+(m-1)d}{a+(m-1)d}\)
\(=\frac{2[a+(m-1)d]}{2[a+(n-1)d]}\)
\(=\frac{2a+(2m-2)d}{2a+(2n-2)d}\)
\(=\frac{2a+(2m-1-1)d}{2a+(2n-1-1)d}\)
\(=\frac{2m-1}{2n-1}\)
[ using (i) by putting m = 2m - 1 and n = 2n - 1 ]

Here, a₁, a₂, a₃, ....., a_n be the given A.P. with a1 as first term and d as the common difference
\(\therefore\)    a₂ = a₁ + d
a₃ = a₁ + 2d
and  a_n = a₁ + ( n - 1 )d
Now,    L.H.S = \(\frac{1}{{a}_{1}{a}_{2}}+\frac{1}{{a}_{2}{a}_{3}}+...+\frac{1}{{a}_{n-1}{a}_{n}}\)
\(=\frac{1}{{a}_{1}({a}_{1}+d)}+\frac{1}{({a}_{1}+d)({a}_{1}+2d)}+...+\frac{1}{[{a}_{1}+(n-2)d][{a}_{1}+(n-1)d]}\)

\(\frac{1}{d}\left[ \frac{1}{{a}_{1}}-\frac{1}{{a}_{1}+d} \right]+\frac{1}{d}\left[ \frac{1}{{a}_{1}+d} -\frac{1}{{a}_{1}+2d}\right]+....+\frac{1}{d}\left[ \frac{1}{{a}_{1}+(n-2)s}-\frac{1}{{a}_{1}+(n-1)d} \right]\)
\(-=\frac{1}{d}\left[ \frac{1}{{a}_{1}}-\frac{1}{{a}_{1}+d}+\frac{1}{{a}_{1}+d}-\frac{1}{{a}_{1}+2d} +....+\frac{1}{{a}_{1}+(n-2)d} -\frac{1}{{a}_{1}+(n-1)d}\right]\)
\(=\frac{1}{d}\left[ \frac{1}{{a}_{1}}-\frac{1}{{a}_{1}+(n-1)d} \right]\)
\(=\frac{1}{d}\left[ \frac{{a}_{1}+(n-1)d-{a}_{1}}{{a}_{1}({a}_{1}+(n-1)d)} \right]\)
\(=\left[ \frac{(n-1)d}{{a}_{1}{a}_{n}} \right]\)
\(=\frac{n-1}{{a}_{1}{a}_{n}}\) = R.H.S.