1. Write the first six terms of an A.P in which a=5,d=4 ?
Ans :- First term , T₁ = a .
          Second term ,T₂ = a+d = 5+4 = 9
         Third term ,T₃ = a+2d = 5+8 = 13
         Fourth term ,T₄ = a+3d = 5+12 = 17
         Fifth term ,T₅ = a+d = 5+16 = 21
         Sixth term ,T₆ = a+d = 5+20 = 25
        ∴ the required A.P is 5,9,13,17,21,25

2. Find the fifth term of 21,28,35,.......
Ans :- From the series ,a=21
                                      d = 28 - 21 = 7.
           ∴ Tn=a+(n-1)d
           ∴ T₅ = a + (5-1) d
                 = a+4d=21+4×7 = 21 + 28 = 49

3. Find the 11 ^th term of the series whose n^th term is 101 - 3n .
 Ans :- Given that T_n = 101 - 3n
                  ∴ T₁₁ = 101 - 3×11
                              = 101 - 33 = 68 .
4. The 5^th term of an AP is 11 and the 9 ^th term is 7.Find the 16^th term .
Ans :- Given that a+4d = 11 and →(1)
                            a+8d = 7  → (2)
       (2) - (1) = 4d = -4
                  ∴ d = -1 .
   from (1) a = 11 - 4d = 11 + 4 = 15
           ∴ T₁₆ = a+15d
                      = 15 + 15×(-1)
                          15 - 15 = 0 .

5. Which term of the series 5,8,11,.....is 320 ?
Ans :- Given that T_n = 320
                     ie,a+(n-1)d = 320
                 Here a = 5 , d = 8 -5 = 3.
               ∴ 5+(n-1)3 = 320 ⇒ 5+3n-3=320
               ie , 3n = 320 - 2 = 318
              ∴  n = ³¹⁸ / ₃ = 106 .

6. The fourth term of an A.P is equal to 3 times the first term , and the seventh term exceeds twice the third term by 1.Find the first term and the common difference.
Ans :- Given that T₄ = 3T₁
                 ie  , a+3d = 3a
                 ⇒ - 2a + 3d = 0 →(1)
   and T₇ = 2T₃ + 1 ⇒ a + 6d = 2(a+2d)+1
         ⇒ -a + 2d = 1 →(2)
    eqn (2) × 2 ⇒ -2a + 4d = 2 →(3)
   (1) - (3) ⇒ -d = -2⇒ d=2
      from (2) ,a=3 .

7. Find the value of K so that 8k+4,6k-2 and 2k+7 will form an A.P?
Ans :- Since the terms are in AP,T₂ - T₁ = T₃ - T₂ .
           ie,6K-2-(8K+4)=2K+7-(6K-2)
            ie,6k-2-8k-4 = 2k+7-6k+2
              ⇒ -2k-6 = -4k+9
               ⇒ 2k=15 ⇒ k=¹⁵/₂ = 7 ¹/₂ .

8. If 7 times the 7^th term of an A.P is equal to 11 times its 11^th term ,show that the 18^th term of the A.P is zero?
Ans :- Given that 7.T₇ = 11 T₁₁
                            ie 7(a+6d)=11(a+10d)
                           ⇒ 7a +42 d = 11a + 110 d
                        ⇒ 4a + 68 d = 0
                       ⇒ a+17 d = 0
                       ⇒ T₁₈  = 0(Since T₁₈ = a+17d)

9. The sums of n terms of two arithmetic series are in the ratio of 2n+1 : 2n-1 .Find the ratio of their 10^th terms ?
Ans :- Let the two A.S be a,a+d,a+2d .......and A,A+D,A+2D.........
      

10. Find the sum of 18 terms of 57 +49+41+ - - -
Ans :- a=57 and d = 49-57=-8 and n=18
∴ S₁₈ = ⁿ/₂[2a+(n-1)d]
   = ¹⁸/₂[2×57+(18-1)-8)
     9[114+17×-8]=9×[114-136]
                                         = - 198

11. Find the sum of numbers between 100 and 200 which are divisible by 7?
Ans :- a = 105 and a_n = 196.
        a_n = 196 ⇒ a+(n-1)d=196
                     105 +(n-1)=196
                   105 + 7n-7=196
                     7n=196+7-105
                               = 98
                             n=⁹⁸/₇ = 14
        ∴ sum = ⁿ/₂ (a₁ + a_n ) = ¹⁴/₂ (105 + 196)=7×301 = 2107

12. The sum of a series of terms in A.P is 128. If the first term is 2 and the last term is 14.Find the common difference ?
Ans :- Given that S= 128
                   ⁿ/₂ (a₁+a_n)=128
           Here a₁ = 2 and an = 14
            ∴ ⁿ/₂ (2+14) = 128 ⇒ n=128ײ/₁₆ = 16
          a_n  = 14 ⇒ a+(n-1)d = 14
                ⇒ 2+15d=14 ⇒ d=¹²/₁₅=⁴/₅ .

13. Find the number of terms of the series 21,18,15,12.........Which must be taken to give a sum of zero?
Ans :- a=21,d=18-21=-3
               Given that sum = 0
                ⇒ ⁿ/₂(2a+(n-1)d)=0
               ⇒ ⁿ/₂[2×21+(n-1)-3]=0
               ⇒ n[42-3n+3]=0⇒ 45=3n⇒n=⁴⁵/₃=15

14. The sum of n terms of a series is (n² + 2n) for all values of n.Find the first 3 terms of the series ?
Ans :- Given that S_n = n² + 2n
                     S_n-1 = (n-1)²+2(n-1)
                             = n² - 2n + 1+2n-2 = n² -1
                       Tn = S_n -S_n-1
                              = n² + 2n - n² + 1 = 2n +1
                       T₁ = 2+1 = 3
                        T₂ = 2×2+1=5
                       T₃ = 2×3+1 = 7 .

15.The sum of the first six terms of an arithmetic progression is 42.The ratio of the 10^th term to the 30^th term of the A.P is 1/3 .Calculate the first term and the 13^th term ?
Ans :- Given that      S₆ = 42 .
 and t₁₀ / t₃₀ = ¹/₃ ⇒ ^a+9d / _a+29d = ¹/₃
                  ⇒ 3a+27d=a+29d
                  ⇒ 2a-2d=0   ⇒ a=d
      S₆ = 42 ⇒ ⁶/₂ (2A+5D)=42
             ⇒ 3(2a+5a)=42⇒21A=42
                              ⇒ a=2
                T₁₃ = a+12d
                          2+12
                 T₁₃ = a+12d
          =2+12×2 = 26

16. If the sums of n,2n,3n terms of an A.S are S₁,S₂ , S₃ respectively .Prove that S₃ = 3(S₂ - S₁ ) ?
Ans :- Given that S_n = S₁
                                  S_2n = S₂
                                   S_3n = S₃
                                    S₁ = ⁿ/₂(2a+(n-1)d)ⁿ/₂[2a+nd-d]
                                  = na+n²d/₂-^nd/₂
                          S₂ = ²ⁿ / ₂ [2a + (2n-1)d]=n[2a+2nd-d]
                                   = 2na + 2n²d-nd
                            S₃ = ³ⁿ/₂[2a+(3n-1)d]=3na+9n²d/₂ - ³/₂ nd.
                 3(S₂-S₁)=3[2na+2n²d-nd-na-¹/₂ n²d+¹/₂ nd]
                               = 3[na+³/₂n²d-¹/₂nd]=3na+⁹/₂n²d-³/₂ nd
                                            =S₃

17.If a,b,c are in A.P Prove that the following are also in A.P
   (i) ¹/_bc , ¹/_ca,¹/_ab   (ii)  b+c , c+a , a+b

Ans ;-(i) a,b,c are in A.P
     Dividing each term by abc,¹/_bc , ¹/_ca , ¹/_ab are in A.P
       (ii) a,b,c   are   in A.P
            substracting a+b+c from each term .
              ⇒ a-(a+b+c), b - (a+b+c) , c - (a+b+c) are in
             ⇒ -(b+c) , -(c+a) , -(a+b) are in A.P
   Multiplying each term by -1, ⇒ b+c , c+a , a+b are in A.P

18. Find the A.M between
   (i)  6 and 12     (ii) 5 and 22
Ans :- (i) A.M between 6 and 12 = ⁶⁺¹²/₂ = ¹⁸/₂ = 9
           (ii) A.M between 5 and 22
                         = ⁵⁺²²/₂ = ²⁷/₂ = 13.5

19. Insert 5 arithmetic means between 4 and 22 ?
Ans :- Let - A₁ , A₂ , A₃ , A₄ , A₅ are the A.M between 4 and 22 .
Then 4,A1 , A2 , A3 , A4 , A5,22 is an A.P containing 7 terms
             22=7^th term ⇒ a+6d = 22
                               ⇒ 4 + 6d = 22
                               ⇒ 6d = 18
                               ⇒  d = 3.
A₁ = 4+3=7
A₂ = 7+3=10
A₃ = 10+3=13
A₄ = 13+3=16
A₅ = 16+3=19.

20.Find two numbers whose product is 91 and whose A.M is 10 ?
Ans :- Let x and y be two numbers.Then
       (x+y)/2=10
       ⇒   x+y = 20 → (1)
      and xy = 91
     (x-y)² = (x+y)² - 4xy
      =  400 - 364
                    = 36.
     ∴ x-y = 6 →(2)
   (1) + (2) ⇒  2x=26
              ⇒  x=13
     from (2) y=x-6
                       = 13-6=7
       Hence the two numbers are 7,13

21. Find the 7 ^th term of 2,4,8,------?
Ans :-   a₁ = 2.
             a₂ = 2×2=4
             a_{3 =} 4×2 = 8
      2,4,8 ,......is a G.P
                 t_n = ar^n-1
                t₇ = ar⁶ = 2×2⁶ = 2⁷ = 128

22.If 5,x,y,z,405 are the first five terms of a G.P,find the values of x,y and z ?
Ans :- Given that t₅ = 405 and a=5
                          ar⁴ = 405 ⇒ 5×r⁴ = 405
                                           ⇒ r⁴ = 81
                                          ⇒ r = 3
  If r = 3 , x = 15 , y = 45 , z = 135
   if r=-3,x=-15,y=-45,z=135

23.The second , third and sixth terms of an A.P are consecutive terms of a geometric progression .Find the common ratio of the geometric progression ?
Ans :- Let a,be the 1^st term of AP and a₂ be the first term of GP.
  then ,a₁ + d = a₂ →(1)
           a₁ + 2d = a₂r →(2)
          a1 + 5d = a₂r²→(3)
   (2) - (1)   ⇒ d=a₂ (r-1)→(4)
   (3) - (2)   ⇒ 3d = a₂r(r-1) →(5)
   ⁽⁵⁾ / ₍₄₎   ⇒ r=3

24.Let A and G be the arithmetic and geometric means of two positive numbers a and b,the A≥G?
Ans :- Proof :-
     Case 1 . If a and b are two unequal positive numbers,
    then A = ^a+b/₂ , G = √ab
  ∴  A - G = ^a+b/₂ - √ab
                 = ^a+b-2√ab/₂
                          = (√a - √b)² / ₂
Since (√a - √b)²  is always +ve ,A-G>0  ⇒ A>G
Case 2 . If a = b , then A = G ,Hence A≥G

25.Find the sum of 8 terms of 3+6+12+ - - - - n?
Ans :- This series is a G.P
  Here r = ⁶/3 = 2
    ∴ S₈ = a (rⁿ-1)/_r-1 = 3(2⁸ - 1)/ _2-1
                              = 3(256 - 1)
                   = 3×255
                  = 765 .

26. Find the sum 1+¹/₂+¹/₄+¹/₈+  - - - ?
Ans :- This is a G.P and we have to find the sum of infinite terms ?
         Here a=1 , r=¹/₂
∴ S_∞=^a / 1-r = ¹/_1-1/2 = ¹/_1/2 = 2

27. Find the sum of n terms of the series 7+77+777+ - - ?
Ans :- Sn = 7+77+777+- - - to nterms
                = 1(1+11+111+ - - - - to n terms )
                 = 7/9 (9+99+999+- - - to n terms )
              = ⁷/₉ {(10-1)+(100-1)+(1000-1)+ - - - to n terms )
                  = ⁷/₉ [{10+10²+10³+ - - - to n terms }-(1+1+1 - - - to n terns }]
                                     = ⁷/₉ [10⁽10ⁿ -1)/_10-1^-n]=7(10ⁿ⁺¹-10/₈₁₎-⁷/₉n.

28. If a,b,c,d are in G.P.P.T a²-b²,b²-c²,c²-d² are also in G.P?
Ans :- Let r be the common ratio of the G.P.Then
              ^b/_a = ^c/_b = ^d/_c = r
               ⇒ b=ar, c=br =ar² , d=cr = ar³
               a²-b² , b²-c² , c²-d² will be in G.P if (b²-c²) = (a²-b²)(c²-d²)
         ie , if (a²r²-a²r⁴)=(a²-a²r²)(a²r²-a²r⁶)
ie if a⁴r⁴(1-r²)2=a²(1-r²)a²r²(1-r²)
                                      = a⁴r⁴(1-r²)²,which is true .
   Hence a²-b², b²-c² , c²-d² are in G.P

29. If m ^th term of an H.P. is n and n ^th term is m,show that r ^th term is ^mn/_r .Hence find (m+n) ^th term ?
Ans :- Let the corresponding A.P be a,a+d,a+2d, - - - -
     Then ,by the question ,m ^th term = ¹/_n
                                 n^th term = ¹/_m
∴ a+(m-1)d = ¹/_n,a+(n-1)d = ¹/_m.
Solving we get a = d = ¹/_mn
          ∴ r th term of the A.P = a+(r-1) d
                                      =¹/_mn + (r-1)¹/_mn = ^r/_mn
∴ r^th term of H.P = ^mn/_r
Put r=m+n .Then (m+n)^th term of the H.P = ^mn/_m+n