Question

1.the 19th term of a.p is equal to 3 times its 6th term if the 9th term of the a.p is 19. Find the first term and it's common difference.
2. If 8 times the 8th term of an a.p is equal to the 12 times the 12th term of a.p then find the 20th term .
3. Explain the reason for the double answer .
4.find a30-a20 for the A. P .5,1,-3,-7........
5.find the number of terms of the a.p 20,16,12,....which when added gives the sum 56. Request for answer as soon as possible....plz ...friends.

Answer 1

Let 'a' be the 1st term(a₁) and 'd' the common difference of the A.P.s.

Ans1=> a₉ = 19.
       a+(9-1)d = 19.
       a+8d=19 .....(1)

 Now, a₁₉ = 3×a₆
         a+(19-1)d=3[a+(6-1)d]
         a+18d=3(a+5d)
         simplifying, we get,
         a=3d/2. ......(2)
Putting value of a in (1)
         3d/2 + 8d=19
Solving we get d=2.
Putting d=2 in (1) we get a=3.                                      (ans.)


Ans2=>
         8×a₈=12×a₁₂
         8[a+(8-1)d]=12[a+(12-1)d]
Solving, we get,
          a = -19d.
          a+19d=0            
          a+(20-1)d=0
          a₂₀=0.                                                            (Ans.)


Ans4=>
           Here, a = 5 and d = -4
           a₃₀ = 5+(30-1)(-4)
                = -111.
           a₂₀ = 5+(20-1)(-4)
                = -71.

a₃₀ - a₂₀ = -111-(-71) = -40.                                       (Ans.)       


Ans5=>

       S=n/2[2a+(n-1)d]
      56=n/2[2×20+(n-1)(-4)]
Solving we get, (n-7)(n-4) = 0.  Hence, n= 7 or 4.
Required no of terms = 4.( sum of 5th term to 7th term is 0). (Ans.)


        

  

Answer 2

T₁₉ = 3 * T₆        and,      T₉ = 19

T₉ = a + (9-1) d = a + 8 d  =  19      --- (1)
T₆ = a + (6-1) d = a + 5 d
T₁₉ =  a + 18 d

 a + 18 d  =  3 * (a + 5 d)  = 3 a + 15 d
    => 2 a - 3d = 0
     =>  a = 3/2 * d      ---(2)
Solving, (1) and(2)  :  a = 3   and d = 2
===================

8 * T₈  = 12 * T₁₂          =>    2  T₈  = 3 T₁₂
=>  2 (a + 7 d)  = 3 (a + 11 d)
=>  a = - 19 d  --- (1)

T₂₀  = a + 19 d =  0      as (1)
===========================
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4.        a  = 5          d =  - 4
   T₃₀ - T₂₀
         = a + 29 d - a - 19 d
         = 10 d   =   - 40
===================
5.   Sum  = S = [ 2 a  + (n--1) d  ] * n / 2
                 56 = [ 2 * 20 - 4 n + 4 ] * n / 2
     =>   28 +  n^2 - 11 n  = 0
     =>   n = (11 + - 3)/2 = 7 or 4

  series      20,  16,  12,  8,  4,  0,  -4,  -8,  -12