6, 9, 12  or  12, 9, 6

173

-176 = 20 + (n - 1) (- 4)  \(\Longrightarrow \)  n = 50 
Then, middle term
= T₂₅ = - 76

407 = 3 + (m - 1) \(\times \) 4 and  709 = 2 + (n - 1) \(\times \) 7
\(\Longrightarrow \) m = 102  and  n = 102
Let pth term of first AP = qth term of second AP.
\(\Longrightarrow \) 3 + (p - 1) \(\times \) 4 = 2 + (q - 1) \(\times \) 7
\(\Longrightarrow \)  \(\frac { p+1 }{ 7 } \) = \(\frac { q }{ 4 } \) = k (say)  \(\Longrightarrow \) p = 7k - 1  and q = 4k
Since, each AP consists of 102 terms.
\(\therefore \)  p \(\le \) 102  and  q \(\le \) 102
\(\Longrightarrow \)  k \(\le \) \(\frac { 103 }{ 7 } \) and k \(\le \) \(\frac { 102 }{ 4 } \) = 14 term

T_z = 7\(\frac { 3 }{ 2 } \)  \(\Longrightarrow \)  a + d = \(\frac { 31 }{ 4 } \)  and T₃₁ = \(\frac { 1 }{ 2 } \)  \(\Longrightarrow \)  a + 30d = \(\frac { 1 }{ 2 } \)
On solving, we get   d = \(\frac { -1 }{ 4 } \) , a= 8
Now, 8 + (n - 1) \(\left( \frac { -1 }{ 4 } \right) \) = \(\frac { -13 }{ 2 } \) = 8, 59

We know the G.M between 'a' and 'b' is \(\sqrt { ab } \)
\(\therefore\)  \(\frac { { a }^{ n+1 }+b^{ n+1 } }{ { a }^{ n }+{ b }^{ n } } ={ a }^{ \frac { 1 }{ 2 } }{ b }^{ \frac { 1 }{ 2 } }\)
\(\Rightarrow\) \({ a }^{ n+1 }+{ b }^{ n+1 }={ a }^{ n+\frac { 1 }{ 2 } }{ b }^{ \frac { 1 }{ 2 } }+{ a }^{ \frac { 1 }{ 2 } }{ b }^{ n+\frac { 1 }{ 2 } }\)
\(\Rightarrow\) \({ a }^{ n+1 }-{ a }^{ n+\frac { 1 }{ 2 } }{ b }^{ \frac { 1 }{ 2 } }={ a }^{ \frac { 1 }{ 2 } }{ b }^{ n+\frac { 1 }{ 2 } }-{ b }^{ n+1 }\)
\(\Rightarrow\) \({ a }^{ n+\frac { 1 }{ 2 } }\left( { a }^{ \frac { 1 }{ 2 } }-{ b }^{ \frac { 1 }{ 2 } } \right) ={ b }^{ n+\frac { 1 }{ 2 } }\left( { a }^{ \frac { 1 }{ 2 } }-{ b }^{ \frac { 1 }{ 2 } } \right) \)
\(\Rightarrow\) \({ a }^{ n+\frac { 1 }{ 2 } }=b^{ n+\frac { 1 }{ 2 } }\)
\(\Rightarrow\) \(\frac { { a }^{ n+\frac { 1 }{ 2 } } }{ { b }^{ n+\frac { 1 }{ 2 } } } =1\)
\(\Rightarrow\) \(\left( \frac { a }{ b } \right) ^{ n+\frac { 1 }{ 2 } }=\left( \frac { a }{ b } \right) ^{ 0 }\)
\(\Rightarrow\) \(n+\frac { 1 }{ 2 } =0\) \(\Rightarrow\) \(n=-\frac { 1 }{ 2 } \)

Let S_n = 8 + 88 + 888 + 8888 + .... upto n terms
= 8 [1 + 11 + 111 + 1111 +.... upto n terms]
= \(\frac { 8 }{ 9 } \) [9 + 99 + 999 + 9999 + ... upto n terms]
= \(\frac { 8 }{ 9 } \) [(10 - 1)+(10² - 1) + (10³ - 1)] +....upto n terms
= \(\frac { 8 }{ 9 } \) [(10 + 10² +10³ + ...n terms) - (1+1+...n terms)]
= \(\frac { 8 }{ 9 } \left[ \frac { 10.\left( { 10 }^{ n }-1 \right) }{ \left( 10-1 \right) } -n \right] \)
= \(\frac { 8 }{ 9 } \left[ \frac { 10 }{ 9 } \left( { 10 }^{ n }-1 \right) -n \right] \)
= \(\frac { 80 }{ 81 } \left[ { 10 }^{ n }-1 \right] -\frac { 8 }{ 9 } n\)

Let the number of sides of polygon be n. The interior angles of the polygon form an A.P.
Here, a = 120° and d=5°
We know that sum of interior angles of a polygon with n sides is (n-2) \(\times\)180°
S_n = (n-2) \(\times\) 180°
\(\Rightarrow \frac{n}{2}[2 a+(n-1) d]=180^{\circ}(n-2) \)
\(\Rightarrow \frac{n}{2}\left[240^{\circ}+(n-1) 5^{\circ}\right]=180(n-2) \)
\(\Rightarrow n[240+(n-1) 5]=360(n-2) \)
\(\Rightarrow 240 n+5 n^{2}-5 n=360 n-720 \)
\(\Rightarrow 5 n^{2}+235 n-360 n+720=0 \)
\(\Rightarrow 5 n^{2}-125 n+720=0 \)
\(\Rightarrow n^{2}-25 n+144=0 \)
\(\Rightarrow n^{2}-16 n-9 n+144=0 \)
\(\Rightarrow n(n-16)-9(n-16)=0 \)
\(\Rightarrow(n-9)(n-16)=0 \)
\(\Rightarrow n=9 \text { or } 16\)

Let 'a' be the first term and 'd' be the common difference of given A.P.
\(\because \quad { S }_{ m }=\frac { m }{ 2 } [2a+(m-1)d]\)
and \({ S }_{ n }=\frac { n }{ 2 } [2a+(n-1)d]\)
\(\therefore \quad \frac { { S }_{ m } }{ { S }_{ n } } =\frac { \frac { m }{ 2 } [2a+(m-1)d] }{ \frac { n }{ 2 } [2a+(n-1)d] } \)
But \(\frac { { S }_{ m } }{ { S }_{ n } } =\frac { { m }_{ 2 } }{ { n }_{ 2 } } \)   [Given]
\(\Rightarrow \frac { \frac { m }{ 2 } [2a+(m-1)d] }{ \frac { n }{ 2 } [2a+(n-1)d] } =\frac { { m }^{ 2 } }{ { n }^{ 2 } } \)
\(\Rightarrow \frac { 2a+(n-1)d }{ 2a+(n-1)d] } =\frac { { m }^{ 2 } }{ { n }^{ 2 } } \times \frac { n }{ 2 } \times \frac { 2 }{ m } =\frac { m }{ n } \)
\(\Rightarrow\) 2an + n(m-1)d = 2am + m(n-1)d
\(\Rightarrow\) 2an-2am= (mn-m)d-(mn-n)d
\(\Rightarrow\) 2a[n-m]=[mn-m-mn+n]d
\(\Rightarrow\) 2a[n-m]=[n-m]d
\(\Rightarrow d=\frac { 2a[n-m] }{ [n-m] } =2a\)
Now, \(\frac { { a }_{ m } }{ { a }_{ n } } =\frac { a+(m-1)d }{ a+(n-1)d } =\frac { a+(m-1)\times 2a }{ a+(n-1)\times 2a } \)
\(=\frac { a[1+2m-2] }{ a[1+2n-2] } =\frac { 2m-1 }{ 2n-1 } \)
Thus, the ratio of 'm^th' and 'n^th' term is (2m+1): (2m-1).

Let A be the first term and d be the common difference of given A.P.
\({ S }_{ p }=\frac { p }{ 2 } [2A+(p-1)d]=a\)
\(\Rightarrow A+\frac { (p-1) }{ 2 } d=\frac { a }{ p } \) ..(i)
\({ S }_{ q }=\frac { q }{ 2 } [2A+(q-1)d]=b\)
\(\Rightarrow \quad A+\frac { (q-1) }{ 2 } d=\frac { b }{ q } \) ...(ii)
\({ S }_{ r }=\frac { r }{ 2 } [2A+(r-1)d]=c\)
\(\Rightarrow \quad A+\frac { (r-1) }{ 2 } d=\frac { c }{ r } \) ....(iii)
Multiplying (i) by (q-r), (ii) by (r-p) and (iii) by (p-q) on both sides and then adding, we have
\(\frac { a }{ p } (q-r)+\frac { b }{ q } (r-p)+\frac { c }{ r } (p-q)\)
\(=\left[ A+\frac { { (p-1) }^{ d } }{ 2 } \right] (q-r)+\left[ A+\frac { { (q-1) }^{ d } }{ 2 } \right] (r-p)+\left[ A+\frac { { (r-1) }^{ d } }{ 2 } \right] (p-q)\)
=A[q-r+r-p+q-q]+d\(\left[ \frac { (p-1) }{ 2 } (q-r)+\frac { (q-1) }{ 2 } (r-p)=\frac { (r-1) }{ 2 } (p-q) \right] \)
\(=A[0]+d\left[ \frac { pq-pr-q+r }{ 2 } +\frac { qr-pq-r+p }{ 2 } +\frac { pr-qr-p+q }{ 2 } \right] \)
= 0 + d\(\left[ \frac { pq-pr-q+r+qr-pq-r+p+pr-qr-p+q }{ 2 } \right] \)
\(=0+d\left[ \frac { 0 }{ 2 } \right] =0+d(0)=0\)

Let a₁,a₂ and d₁,d₂ be the first terms and common differences of two A.P's respectively.
\(\therefore { S }_{ n }=\frac { n }{ 2 } [2{ a }_{ 1 }+(n-1){ d }_{ 1 }]\)
and \({ S }_{ n }^{ ' }=\frac { n }{ 2 } [2{ a }_{ 2 }+(n-1){ d }_{ 2 }]\)
Now \(\frac { { S }_{ n } }{ { S' }_{ n } } =\frac { \frac { n }{ 2 } [{ 2a }_{ 1 }+(n-1){ d }_{ 1 }] }{ \frac { n }{ 2 } [{ 2a }_{ 2 }+(n-1){ d }_{ 2 }] } \)
\(=\frac { 2{ a }_{ 1 }+(n-1){ d }_{ 1 } }{ { 2a }_{ 2 }+(n-1){ d }_{ 2 } } \)
But \(\frac { { S }_{ n } }{ { S }_{ n }^{ ' } } =\frac { 5n+4 }{ 9n+6 } \)   [Given]
\(\therefore \frac { 5n+4 }{ 9n+6 } =\frac { { 2a }_{ 1 }+(n-1){ d }_{ 1 } }{ { 2a }_{ 2 }+(n-1){ d }_{ 2 } } \)
\(=\frac { { a }_{ 1 }+\left( \frac { n-1 }{ 2 } \right) { d }_{ 1 } }{ { a }_{ 2 }+\left( \frac { n-1 }{ 2 } \right) { d }_{ 2 } } \)
Now to get 18^th terms \(\frac { n-1 }{ 2 } =17\)
\(\therefore\) n = 35
Putting n = 35
\(\therefore \frac { 5\times 35+4 }{ 9\times 35+6 } =\frac { 2{ a }_{ 1 }+(35-1){ d }_{ 1 } }{ { 2a }_{ 2 }+(35-1){ d }_{ 2 } } \)
\(\Rightarrow \frac { 179 }{ 321 } =\frac { 2({ a }_{ 1 }+17{ d }_{ 1 }) }{ 2({ a }_{ 2 }+17{ d }_{ 2 }) } \)
\(\Rightarrow \frac { { a }_{ 1 }+17{ d }_{ 1 } }{ { a }_{ 2 }+17{ d }_{ 2 } } =\frac { 179 }{ 321 } \)
\(\Rightarrow \frac { { a }_{ 1 }+17{ d }_{ 1 } }{ { a }_{ 2 }+17{ d }_{ 2 } } =\frac { 179 }{ 321 } \)
Thus the ratio of 18th terms of two A.P's is 179:321