Question

1.For what value of n, the nth term of the two APs 1,7,13,19.....And 64,63,62...are same?
2.which term of the sequence 48,43,38,33... is the first negative term?
3.The angles of a quadrilateral are in A.P and the greatest angle is thrice the least angle.find the angles of the quadrilateral?
4.Find the S.I on rs 100000 @8% p.a at the end of each year. also find by using the formula to find the nth term of an A.P., the interest at the end of the 20 years.
5.The last term of an A.P is 120. its first term and common difference are 20 and 5 respectively.find the sum of all terms of A.P.
6.If the 19th term of an a.P. is 39,find the sum of its 37 terms.
7.How many terms of the A.P.2,4,6,.... are needed to give the sum 132?
8. find the sum of the first 24 terms of an A.P. whose nth term is given a n=3+2n.
9.Find the A.P.whose sum to n term is 4nsquare -n.
10. The sum of n term of an A.P. is n(5n-3). find the A.P.
hey friendz ... plz help me out for this questions....:)

Answer 1

1, 7, 13, 19,...            T_n = 1 + 6 n
64, 63, 62, ...             T_n = 64 - n * 1
         1 + 6 n = 64 - n        =>        n = 9

================================
     T_n = 48 - 5 n      = 0 for  n = 9.6
   So T_n < 0  for    n = 10

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A + B + C + D  = 360
   =>  [2 A + (4 - 1) d ] * 4/2 = 360
   =>  2 A + 3 d = 180    --- (1)

 and,  D = 3 A  =>    A + 3 d = 3 A
           =>  2 A = 3 d  --- (2)

  Solving,  A = 45 deg.
                 d = 30 deg.
======================
S.I.  =  Rs 1,00,000 @ 8% p a.
       = 1,00,000 * 8/100 * 1  = Rs 8, 000 per year
it is not clear ho w    AP  series can be used here.

Simple interest at the end of 20 years :  20 * 8, 000 = Rs 1, 60, 000
=======================
T_n = 120
20 + (n - 1) 5 = 120    =>        n = 21
S_n =  [2 a + (n-1)d ] n/2  =  [ 40 + 20*5 ] * 21/2  =  1470
===========
T_19 = 39
a + 18 d = 39
S_37  = [2 a + 36 d ] * 3 7 /2  =  (a+18d) 37 = 39 * 37 = 1443
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[2*2 + (N - 1) 2 ]  N /2  = 132
   =>   N^2 +  N -  132  = 0
   =>  N = [- 1 + - √529 ] / 2
            = 11  or  -12
==============
  n = 24           
Tn =  3  + 2 n    =>    a = 3 +2 = 5
d = T2 - T1 = 7 - 5 = 2

Sn = [ 2 * 5 + 23 * 2 ] * 24/2       = 672
==============
Sn =  [ 2 a + (n - 1) d ] * n / 2  = 4 n^2 -  n
   =>    a  n  + d n^2 /2 - d n / 2  = 4 n^2  - n
     =>  a - d/2  =  -1
     =>  d / 2  = 4
                     =>so  d = 8      and      a = 3
======================
     Sn = [ 2 a  + (n - 1) d ] n / 2  =  5  n^2  - 3  n
       =>      a n + n^2 d /2  - n d / 2  =  5  n^2  - 3  n
       =>      d/2 = 5        =>  d = 10
       =>      a - d/2 = -3
           =>  a = 2