\(S_{n} =5 n^{2}+3 n \)
\(S_{n-1} =5(n-1)^{2}+3(n-1) \)
\(=5\left(n^{2}-2 n+1\right)+3(n-1) \)
\(=5 n^{2}-10 n+5+3 n-3 \)
\(=5 n^{2}-7 n+2 \)
\(\text { Now } n^{\text {th }} \text { term }=S_{n}-S_{n-1}\)
\(\therefore \text { The required } n^{\text {th }} \text { term }=\left[5 n^{2}+3 n\right]-\left[5 n^{2}-7 n+2\right]\)
\(=5 n^{2}+3 n-5 n^{2}+7 n-2\)
\(t_{n}=10 n-2\)

\(a=-1, d=t_{2}-t_{1}=3-(-1)=4\)
\(\text { Let } 95 \text { be the } n^{\text {th }} \text { term of the A.P. }\)
\(\text {then } t_{n}=95\)
\(a+(n-1) d =95 \)
\(-1+(n-1) 4 =95 \)
\((n-1) 4 =95+1=96 \)
\(n-1 =\frac{96}{4}=24\)
\(n=24+1=25\)
\(\therefore 95 \text { is the } 25^{\text {th }} \text { term of the A.P. }\)

Let the four terms be (a - 3d),(a - d) (a + d), (a + 3d)
Sum = 20 (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20
a - 3d + a - d + a + d + a + 3d = 20
4a = 20
\(a=\frac{20}{4}=5\)
Given sum of squares = 120
\((a-3 d)^{2}+(a-d)^{2}+(a+d)^{2}+ (a+3 d)^{2}=120 \)
\(a^{2}+9 d^{2}-6 a d+a^{2}+d^{2}-2 a d+a^{2}+d^{2}+2 a d+a^{2}+9 d^{2}+6 a d=120\)
\(4 a^{2}+20 d^{2} =120 \)
\(a^{2}+5 d^{2} =30 \)
\(25+5 d^{2} =30 \)
\(5 d^{2} =5 \)
\(d =\pm 1 \)
If d= 1 then the number are 2,4,6,8
If d= -1 then the number are 8,6,4,2
therefore the four numbers are 2,4,6,8

\(\text { Given } 2 \mathrm{x}, \mathrm{x}+10,3 \mathrm{x}+2 \text { are in A.P }\)
\(x+10-2 x =3 x+2-[x+10] \)
\(-x+10 =3 x+2-x-10=2 x-8 \)
\(10+8 =2 x+x \)
\(3 x =18 \)
\(x =\frac{18}{3}=6\)

Given \(a_{n} =\frac{2 n-3}{6} \)
\(\therefore a_{1} =\frac{2(1)-3}{6}=\frac{2-3}{6}=\frac{-1}{6} \)
\(a_{2} =\frac{2(2)-3}{6}=\frac{4-3}{6}=\frac{1}{6} \)
\(a_{3} =\frac{2(3)-3}{6}=\frac{6-3}{6}=\frac{3}{6} \)
\(\therefore \text { First three terms are } \frac{-1}{6}, \frac{1}{6}, \frac{3}{6} \text {. }\)

Given the sequence 5, 8, 11,... Every term is 3 more than the previous term but the first term is 5.
The general term may be a_n = 3n + 2

In each term the numerator increases by n (n - 2) and denominator is 2.
The general term is \(a_{n}=\frac{n(n-2)}{2}\)

Given \(a_{n} =\frac{n(n-3)}{n+4} \)
\(a_{15} =\frac{15(15-3)}{15+4}=\frac{15 \times 12}{19} \)
\(a_{15} =\frac{180}{19} \)

Given \(a_{n}=(-1)^{n-1} 2^{n}\)
\(\mathrm{a}_{1} =(-1)^{1-1} 2^{1} \)
\(=(-1)^{0} 2^{1}=1 \times 2=2 \)
\(\mathrm{a}_{2} =(-1)^{2-1} \times 2^{2} \)
\(=(-1)^{1} \times 4=-4 \)
\(\mathrm{a}_{3} =(-1)^{3-1} \times 2^{3} \)
\(=(-1)^{2} \times 8=8\)
\(\therefore \text { First three terms are } 2,-4,8\)

\(3^{2} =1(\bmod 8) \)
\(\left(3^{2}\right)^{101} \equiv 1^{101}(\bmod 8) \)
\(3^{202} \equiv 1(\bmod 8) \)
\(5^{2} \equiv 1(\bmod 8) \)
\(5 \equiv 5(\bmod 8) \)
\(\left(5^{2}\right)^{4} \equiv 1^{4}(\bmod 8) \)
\(5^{8} .5^{1} \equiv 5(\bmod 8) \)
\(3^{202}+5^{9} =6(\bmod 8) \)
Remainder is 6 when divided by 8

We have
\(2^{3} \equiv 1(\bmod 7) \)
\(\left(2^{3}\right)^{8} \equiv 1^{8}(\bmod 7) \)
\(2^{3} 2^{24} \equiv 1 \times 1^{8}(\bmod 7) \)
\(2^{27} \equiv 1(\bmod 7) \)
\(2^{2} 2^{27} \equiv 4 \times 1(1 \bmod 7) \)
\(2^{29} \equiv 4(\bmod 7) \)
\(2^{29}+3 =(4+3)(\bmod 7) \)
\(2^{29}+3 \equiv 0(\bmod 7) \)
\(2^{29}+3 \text { is divisible by } 7\)

\(\text { Interval of change }=\text { L.C.M. of }(48,72,108) \text { sec }\)
\(\begin{aligned} &\begin{array}{l|l} 2 & 48,72,108 \\ \hline 3 & 4,6, 9 \\ \hline 2 & 4,2, 3 \\ \hline & 2,1, 3 \end{array} \end{aligned}\)
LCM = 12 x 3 x 2 x 2 x 3 = 432
So the lights will again change simultaneously after every 432 sec
= 432/60 = 7 min 12 sec
Hence next change will be at 8 : 27 : 12 am.

Smallest composite number = 4 = 2²
Smallest prime number = 2 = 2¹
Therefore H.C.F of 4 and 2 = 2

By fundamental theorem we have every composite number can be expressed as a product of primes
Factorizing 100 and 190

\(100=2^{2} \times 5^{2} \)
\(190=2^{1} \times 5^{1} \times 19^{1} \)
\(\therefore H C F \text { of } 100 \text { and } 190=2^{1} \times 5^{1}= 10\)
\(\text { H.C.F. } \times \text { L.C.M. }=\text { Product of two numbers }\)
\(10 \times \text { L.C.M. }=100 \times 190\)
\(\text { L.C.M. }=\frac{100 \times 190}{10}=1900\)

We have 6ⁿ = (2 x 3 )ⁿ
= 2ⁿ x 3ⁿ
prime factorization of 6ⁿ does not contain the prime number 7. Therefore 6ⁿ is not divisible by 7