Three digit numbers which are divisible by 7 are 105,112,119, ....,994.
It is in A.P. where a = 105, d = 7
\(t_{n} =994=l \)
\(t_{n} =a+(n-1) d \)
\(994 =105+(n-1) 7 \)
\(n-1 =\frac{994-105}{7}=\frac{889}{7}=127 \)
\(n =127+1=128 \)
\(\text { Now } \mathrm{S}_{\mathrm{n}}=\frac{n}{2}(a+l)\)
\(=\frac{128}{2}(105+994)\)
\(=64(1099)\)
\(\therefore \text { The required sum }=70336\)

First term a = 17
Last term l = 350 = t_n
Common difference d = 9
\(t_{n}=a+(n-1) d\)
\(350=17+(n-1) 9\)
\(\frac{350-17}{9}=n-1\)
\(n=\frac{333}{9}+1\)
\(=37+1=38\)
\(\therefore \text { There are } 38 \text { terms in the A.P. }\)
\(\mathrm{S}_{\mathrm{n}}=\frac{n}{2}(a+l) \)
\(\mathrm{S}_{38}=\frac{38}{2}(17+350) \)
\(=19 \times 367\)
\(\therefore \text { Required sum }=6973\)

We have a = 5 ;d = 10
\(a_{31}=a+30 d=5+30 \times 10=305\)
Let the n^th term of the Given A.P. is 130 more than the 31^th term.
\(a_{n} =130+a_{31} \)
\(a+(n-1) d =130+305 \)
\(5+10(n-1) =435 \)
\(10(n-1) =430 \)
\(n-1 =43 \)
\(n =44 \)
Therefore 44^th term of the given A.P. is 130 more than it's 31^th term.

Let a be the first term and d be the common difference of the A.P.
\(\text {Given } \mathrm{a}_{8}=31 \text { and } a_{15}=16+a_{11} \)
\(a+(n-1) d =t_{n} \)
\(a+7 d =31 \text { and } \)
\(a+14 d =16+(a+10 d) \)
\(a+14 d =16+a+10 d \)
\(a-a+14 d-10 d =16 \)
\(4 d =16 \)
\(d=\frac{16}{4}=4\)
\(\text {Taking }a+7 d=31 \)
a + 7 (4) = 31
a + 28 = 31
a = 31 -28 = 3.
\(\therefore \text { The A.P. is } \mathrm{a}, \mathrm{a}+\mathrm{d}, \mathrm{a}+2 \mathrm{~d}, \ldots\)
\(\Rightarrow 3,7,11,15,19, \ldots\)

Consider the A.P. 69, 68, 67,...
\(a =69 ; d=68-69=-1 \)
\(t_{n} =a+(n-1) d \)
\(t_{n} =69+(n-1)(-1) \)
\(\text {For the A.P. } 1,7,13,19, \ldots\)
\(a =1 ; d=7-1=6 \)
\(T_{n} =1+(n-1) 6 \)
\(\text {If the two A.P's has an identical term then } t_{n}=T_{n} \text { for some } n \text {. }\)
\(69+(n-1)(-1) =1+(n-1) 6 \)
\(69-n+1 =1+6 n-6 \)
\(69+1-1+6 =6 n+n \)
\(75 =7 n \)
\(n=\frac{75}{7}\)
which is not a natural number.
The two A.P's do not have an identical term for any n

\(\text{Given } a_{n}=(n-1)(2-n)(3+n)\)
\(\text { Put } \mathrm{n}=1 ; \ \mathrm{a}_{1}=(1-1)(2-1)(3+1)\)
\(a_{1}=0 \times 1 \times 4=0\)
\(\text { Put } \mathrm{n}=2 ; \ \mathrm{a}_{2}=(2-1)(2-2)(3+2)\)
\(a_{2}=1 \times(0) \times 5=0\)
\(\text { Put } n=3 ; \ a_{3}=(3-1)(2-3)(3+3)\)
\(=2 \times(-1) \times 6=-2 \times 6\)
\(a_{3}=-12\)
\(a_{1}=0 ; a_{2}=0 ; a_{3}=-12\)

\(\text { Given } a_{n}=(-1)^{n} n\)
\(\text { We know that }\)
\((-1)^{\text {odd number }}=\text { '_ } ^{\text {' }} \text { ve }\)
\((-1)^{\text {even number }}=+\mathrm{ve}\)
\(a_{3}=(-1)^{3} 3=-3 \)
\(a_{5}=(-1)^{5} 5=-5 \)
\(a_{8}=(-1)^{8} 8=8 \)
\(a_{3}=-3 ; a_{5}=-5 ; a_{8}=8 \)

\(\text { Given } a_{1} =3 . \)
\(a_{n} =3 a_{n-1}+1 \text { for all } n>1 \)
\(\text { Put } n =2 ; \)
\(a_{2} =3 a_{2-1}+1=3 a_{1}+1 \)
\(=3(3)+1=9+1=10 \)
\(\text { Put } n =3 ; \)
\(a_{3} =3 a_{3-1}+1=3 a_{2}+1 \)
\(=3(10)+1=30+1=31 \)
First three terms are 3, 10, 31.

Monday stands for 1.
29 days after Monday is 1 + 29 (mod7)
\(\equiv\) 30 (mod 7)
\(\equiv\) 2 (mod 7)
2 stands for Tuesday.
Results will be published on Tuesday.

\(\begin{array}{l|l} 2 & 12,18,21,30 \\ \hline 3 & 6,9,21,15 \\ \hline &2,3, 7,5 \end{array}\)
\(\text { I.C.M of } 12.18,21,30=2^{2} \times 3^{2} \times 5^{1} \times 7^{1}\)
= 4 x 9 x 35
= 1260
Required number = 1260 \(\div\) 2
= 630

Required number = H.C.F of (91 - 43), (183 - 91) and(183 -43)
= H.C.F. of 48, 92,and 140

\(48 =2^{4} \times 3^{1} \)
\(92 =2^{2} \times 23^{1} \)
\(140 =2^{2} \times 5^{1} \times 7^{1} \)
\(\therefore H.C.F. is \ 2^{2}=4\)
\(\therefore \text { The required number is } 4\)

Required Length = H.C.F. of (7 m,3 m, 85 cm,12 m 95 cm)
= H.C.F. of 700 cm,385 cm,1295 cm

\(700 =2^{2} \times 5^{2} \times 7^{1} \)
\(385 =5^{1} \times 7^{1} \times 11^{1} \)
\(1295 =5^{1} \times 7^{1} \times 37^{1} \)
\(\text { H.C.E is } 5^{1} \times 7^{1}=35\)
\(\therefore \text { The required length is } 35 \mathrm{~cm} \text {. }\)

a_n = 3 +2n
a₁ = 3 + 2 = 5
a₂ = 3 + 2 x 2 = 7
a₃ = 3 + 2 x 3 = 9
List of numbers becomes 5, 7, 9, 11,.........
Here, 7 - 5 = 9 - 7 = 11 - 9 = 2 and so on. So, it forms an A.P. with common difference d = 2.
To find S₂₄ we have n = 24, a = 5, d = 2.
S₂₄ = \(\frac{24}{2}\) [2 x 5 + (24 - 1) x ]
= 12 [10 + 46] = 672.
So, sum of first 24 terms of the list of numbers is 672.

Here a = 24, d = 21-24 = -3, S_n = 78. We need to find n.
We know that,
S_n= \(\frac{n}{2}\) (2a + (n - 1)d)
78 = \(\frac{n}{2}\) 48 + (n -1)(-3))
78 = \(\frac{n}{2}\) 2(51-3n)
or 3n² - 51n + 156 = 0
n² -17n + 52 = 0
(n - 4)(n - 13) = 0
n = 4 or 13
The number of terms are 4 or 13.

Here S₁₄ = 1050
n = 14
a = 10
S_n = \(\frac{n}{2}\) 2(2a + (n -1)d)
1050 = \(\frac{14}{2}\)(20 + 13d)
= 140 + 91d
910 = 91d
d = 10
a₂₀ = 10 + (20 - 1) x 10
= 20
∴ 20th term = 200.