\(\therefore\) A.M. of a and b = \(\frac{a+b}{2}\)
G.M of a and b = \(\sqrt{ab}\)
\(\therefore\)  \(\frac{a+b}{2\sqrt{ab}}=\frac{m}{n}\)
By componendo and dividendo, we get
\(\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{m+n}{m-n}\)
\(\Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{m+n}{m-n}\)
\(\Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{m+n}}{\sqrt{m-n}}\)
Again by componendo and dividendo, we have:
\(\frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}\)
\(\Rightarrow \frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}\)
Squaring both sides,
\(\frac{a}{b}=\frac{(\sqrt{m+n}+\sqrt{m-n})^2}{(\sqrt{m+n}-\sqrt{m-n})^2}\)
\(\Rightarrow \frac{a}{b}=\frac{m+n+m-n+2\sqrt{(m+n)+\sqrt{m-n}}}{m+n+m-n-2\sqrt{(m+n)(m-n)}}\)
\(\Rightarrow \frac{2m+2\sqrt{m^2-n^2}}{2m-2\sqrt{m^2-n^2}}\)
\(\Rightarrow \frac{a}{b}=\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}\)
Thus, a: b = \((m+\sqrt{m^2-n^2}):(m-\sqrt{m^2-n^2})\)

Let a be the first term and r the common ratio of the given series 
Then (a+ ar + ar² + ar³....\(\infty\)) = 15
\(\frac { a }{ 1-r } =15\) ......(i)
and (a² +a²r² +a²r⁴ +a²r⁶ +....\(\infty\)) = 45
\(\frac { { a }^{ 2 } }{ 1-{ r }^{ 2 } } =45\) .....(ii)
Squaring (l) both sides, we get
\(\Rightarrow\)  \(\frac { { a }^{ 2 } }{ \left( 1-r \right) ^{ 2 } } =225\) .....(iii)
\(\frac { \frac { { a }^{ 2 } }{ \left( 1-r \right) ^{ 2 } } }{ \frac { { a }^{ 2 } }{ 1-{ r }^{ 2 } } } =\frac { 225 }{ 45 } \)
\(\Rightarrow\)\(\frac { 1-{ r }^{ 2 } }{ \left( 1-r \right) ^{ 2 } } =5\)
\(\Rightarrow\) \(\frac { 1+r }{ 1-r } =5\)
\(\Rightarrow\) 1 + r = 5 - 5r \(\Rightarrow\) 6r = 4
r  = \(\frac { 2 }{ 3 } \)
\(\therefore\) Putting the value of r is equation (i) we get
\(\frac { a }{ 1-\frac { 2 }{ 3 } } =15\)
\(\therefore\)  a = 5
\(\therefore\) The required series is \(\left( 5+\frac { 10 }{ 3 } +\frac { 20 }{ 9 } +\frac { 40 }{ 27 } +...\infty \right) \)

Since a, b, c are in AP.
b - a = c - b \(\Rightarrow\) 2b = a + c ..(i)
Now \({ A }_{ 1 }=\frac { 1 }{ 2 } \left( a+b \right) \)
and \(A_{ 2 }=\frac { 1 }{ 2 } (b+c)\)
\(\therefore\)  A.M of A₁ and A₂ = \(\frac { 1 }{ 2 }\) (A₁ + A₂)
= \(\frac { 1 }{ 2 } \left[ \frac { 1 }{ 2 } \left( a+b \right) +\frac { 1 }{ 2 } \left( b+c \right) \right] \)
= \(\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \left[ a+2b+c \right] \)
= \(\frac { 1 }{ 4 } [2b+2b]\left[ \because \quad a+c=2b \right] \)
= \(\frac { 1 }{ 4 } \times 4b=b\)

The given series is 3 x 1² + 5 x 2² + 7 x 3² + ... + to n terms.
Let 'a_n' be the nth term of given series and
'S_n' be the sum of n terms.
\(\therefore\) a_n = [n^th term of 3, 5, 7 ...] [n^th term of 1, 2, 3, ....]²
= (2n + 1) (n)² = 2n³ + n²
\(\therefore { S }_{ n }=\overset { n }{ \underset { k=1 }{ \Sigma } } { a }_{ k }=\overset { n }{ \underset { k=1 }{ \Sigma } } \left( { 2k }^{ 2 }+{ k }^{ 2 } \right) \)
= (2· 1³ + 1²)+ (2· 2³ + 2²)+ (2· 3³ + 3²) + ... + (2. n³ + n²)
= 2(1³ + 2³ + 3³ + ... + n³) + (1² + 2² + 3² + ... + n²)
= \(2[\frac{n(n+1)}{2}]^2+\frac{n(n+1)(2n+1)}{6}\)
= \(\frac{2n(n+1)^2}{4}+\frac{n(n+1)(2n+1)}{6}\)
= \(\frac{n(n+1)}{2}[n(n+1)+\frac{(2n+1)}{3}]\)
= \(\frac{n(n+1)}{2}[\frac{3(n^2+n)+(2n+1)}{3}]\)
= \(\frac{n(n+1)}{2}[\frac{3n^2+3n+2n+1}{3}]\)
= \(\frac{n(n+1)(3n^2+3n+2n+1)}{6}\)
= \(\frac{n(n+1)(3n^2+5n+1)}{6}\)

The given series is 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... to n terms.
Let an be the nth term of the given series.
a = [n^th term of 1, 2, 3, ...] [n^th term of 2, 3, 4, 5...]
= [1 + (n - 1) x 1] [2 + (n - 1) x 1]
= n(n + 1) = n² + n
\(\therefore { S }_{ n }=\overset { n }{ \underset { k=1 }{ \Sigma } } { a }_{ k }=\overset { n }{ \underset { k=1 }{ \Sigma } } \left( { k }^{ 2 }+k \right) \)
= [1² + 1] + [2² + 2] + [3² + 3] + ... + [n² + n]
= [1² + 2² + 3² + ... + n²] + [ 1 + 2 + 3 + ... + n]
= \(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\)
\(\Rightarrow \frac{n(n+1)}{2}[\frac{2n+1}{3}+1]\)
= \(\frac{n(n+1)}{2}\times \frac{(2n+4)}{3}\)
\(\Rightarrow \frac{n(n+1)(n+2)}{3}\)

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.
Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d
Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d]+ [a + n – 1) d]
= 4a + (4n – 10) d
According to the given condition,
4a + 6d = 56
⇒ 4(11) + 6d = 56 [Since a = 11 (given)]
⇒ 6d = 12
⇒ d = 2
∴ 4a + (4n –10) d = 112
⇒ 4(11) + (4n – 10)2 = 112
⇒ (4n – 10)2 = 68
⇒ 4n – 10 = 34
⇒ 4n = 44
⇒ n = 11
Thus, the number of terms of the A.P. is 11.

Let 'r' be the common ratio for the given G.P.
Here, a = 1 and a₃ + a₅ = 90
\(\therefore\)  ar² + ar⁴ = 90 \(\Rightarrow\) a (r² + r⁴) = 90
\(\therefore\)  r² + r⁴ = 90  \(\Rightarrow\)  r⁴ +r² -90 = 0
which is a quadratic equation in r²
\(\therefore\) r² = \(\frac { -1\pm \sqrt { \left( 3 \right) ^{ 2 }-4\times (-90)\times 1 } }{ 2\times 1 } \)
\(\Rightarrow\) r² = \(\frac { -1\pm \sqrt { 1+360 } }{ 2 } =\frac { -1\pm \sqrt { 361 } }{ 2 } \)
r²  = \(\frac { -1\pm 19 }{ 2 } \)
Either r² = \(\frac { -1\pm 19 }{ 2 } \) i.e r² = \(\frac { 18 }{ 2 } \)
\(\Rightarrow\) r² = 9  r = \(\pm\) 3
or  r² = \(\frac { -1-19 }{ 2 } \) i.e r² = \(\frac { -20 }{ 2 } \)
\(\Rightarrow\) r² = -10, which is not possible

Let 'a' be the 1^st term and 'd' be the common difference of the given AP.
\(\therefore\) S_n = \(\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(\therefore\) S_n = \(\frac { 2n }{ 2 } \left[ 2a+\left( 2n-1 \right) d \right] \)
\(\therefore\)  S_n = \(\frac { 3n }{ 2 } \left[ 2a+(3n-1)d \right] \)
Now S₂ - S₁ = \(\frac { 2n }{ 2 } \left[ 2a+(2n-1)d \right] -\frac { n }{ 2 } \left[ 2a+(n-1)d \right] \)
= \(\frac { n }{ 2 } \left[ 4a+4nd-2d-2a-nd+d \right] \)
= \(\frac { n }{ 2 } \left[ 2a+3nd-d \right] \)
= \(\frac { n }{ 2 } [2a+(3n-1)d]\)
3(S₂-S₁) = \(\frac { 3n }{ 2 } [2a+(3a-1)d]\) = S₃
[\(\therefore\) S₃ = \([2a+(3a-1)d]\)
Thus, S₃ = 3 (S₂-S₁)

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 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 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

Let a be the first term and d be the common difference of given A.P.
Here \({ a }_{ p }=\frac { 1 }{ q } \) and \({ a }_{ q }=\frac { 1 }{ p } \)
\(\therefore \quad a+(p-1)d=\frac { 1 }{ q } anda+(q-1)d=\frac { 1 }{ p } \)
\(a+pd-d=\frac { 1 }{ q } \)        .....(i)
and \(a+qd-d=\frac { 1 }{ p } \)  .....(ii)
Subtracting (ii) from (i)
a + pd - d - (a + qd-d) = \(\frac { 1 }{ q } -\frac { 1 }{ p } \)
\(\Rightarrow\) a + pd - d - a - qd+d = \(\frac { p-q }{ pq } \)
\(\Rightarrow \quad (p-q)d=\frac { p-q }{ pq } \)
\(\Rightarrow \quad d=\frac { p-q }{ pq } \times \frac { 1 }{ p-q } \)
\(=\frac { 1 }{ pq } \)
Putting value of d in (i)
a + (p-1) \(\times \frac { 1 }{ pq } =\frac { 1 }{ q } \)
\(\Rightarrow a=\frac { 1 }{ q } -\frac { p-1 }{ pq } =\frac { p-p+1 }{ pq } =\frac { 1 }{ pq } \)
Now \({ S }_{ n }=\frac { n }{ 2 } [2a+(n-1)d]\)
\(\therefore \quad { S }_{ pq }=\frac { pq }{ 2 } \left[ 2\times \frac { 1 }{ pq } +(pq-1)\times \frac { 1 }{ pq } \right] \)
\(=\frac { pq }{ 2 } \left[ \frac { 2 }{ pq } +\frac { pq-1 }{ pq } \right] =\frac { pq }{ 2 } \left[ \frac { 2+pq-1 }{ pq } \right] \quad \)
\(=\frac { pq }{ 2 } \left[ \frac { pq+1 }{ pq } \right] =\frac { pq+1 }{ 2 } \)
\(\therefore \quad { S }_{ pq }=\frac { 1 }{ 2 } (pq+1)\)