the sum of first six terms of an ap is 42 .the ratio of its 10th term is and its 30th term is 1:3. calculate the first and thirteenth term

= n/2 [ 2a + (n-1)d ] = 3[2a+5d]= 6a+15d = 2a+5d..... = 14 ... (EQN-1)

a+9d/ a+29d = 1/3

3a+27d = a+29d

2a-2d ..............(EQN-2)

From eqn 1 -2

2a+5d= 14

2a-2d  ( onsubtracting)

---------------

  7d=14

  d= 2

--------------

putting value of d in eqn (2)

2a-2d

2a -4=0

2a=4

a=2 (ANS)

13term = a+12d= 2+24= 26 (ANS)

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Easy :

As usual, Let the first term be a and the common difference be d

thus, A to Q,

S6 = 42

=> 6/2 (2a + 5d) = 42

=> 3 (2a + 5d) = 42

=> 2a + 5d = 14 ----- (i)

Again, T10/T30 = 1 : 3

=> (a + 9d)/(a + 29d) = 1/3

=>a + 29d = 3a + 27d

=> 2a - 2d = 0

=> a - d = 0 --------- (ii)

On solving (i) and (ii) we get,

a = 2 and d = 2

Thus, first term = 2 and Thirteenth term = 2 + 12(2) = 2 + 24 = 26 Ans....!!!!

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I didn't think someone was even quicker....!!!!

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well that was me :D

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@Akansha, Rahul is just trying to be like his account photo, JB!!! ;P
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Here is the solution = n/2 [ 2a + (n-1)d ] = 3[2a+5d]= 6a+15d = 2a+5d..... = 14 ... (EQN-1) a+9d/ a+29d = 1/3 3a+27d = a+29d 2a-2d ..............(EQN-2) From eqn 1 -2 2a+5d= 14 2a-2d ( onsubtracting) --------------- 7d=14 d= 2 -------------- putting value of d in eqn (2) 2a-2d 2a -4=0 2a=4 a=2 (ANS) 13term = a+12d= 2+24= 26 (ANS)
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26
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18th term is 38
 
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26
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t10/t30=1/3
a+9d/a+29d=1/3
3a+27d=a+29d
2a=2d
a=d
S6=n/2(2a+n-1)d
42 =3(2a+5a)
42=21a
a=2
a13=a+12d
      =2+24
      =28
  • -12
wow
 
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Hey rahul mishra you answered the ques. Very well... It is very usefull...thanks for it😘😘..
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3 seats of English , Hindi and mathematics books have to be stacked in such a way that all the books are stored topic wise and height of each stack is the same.The number of English book is 96, the number of Hindi book is 240 and the number of Mathematic book is 336.Assuming that the books are of the same thickness determine the number of stacks of English , hindi and mathematics.
  • -6
Show that sum of (m+n)th and (m-n)th term of an AP id equal to twice the mth term Find fast please
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yes
 
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S6 =42

a + 9d           1
                 =         
a + 29d          3

cross multiply we  get

3a + 27d = a +29 d

2a - 2d = 0 (1)
 
its given that 
sum of first six terms of an AP is 42

therefore 

S6 = n/2 ( 2a + (n-1) d)

42 = 6/2 ( 2a + (6-1) d)

42 = 3 (2a + 5d )

14 = 2a +5d

2a +5d = 14  (2)

solve eq 1 and 2

2a - 2d = 0
2a +5d = 14

we get 
d= 2
a = 2

13th term of AP
= a + (n-1)d

2+ (13-1) 2

= 2+ 24

=26
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This is the answer

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From which book is this question taken from
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s6 = 42
by the formula Sn=n/2{2a+(n-1)d}
substituting the values
6/2[2a+(6-1)d]=42
3[2a+5d]=42
2a+5d=42/3
2a+5d=14---------------(1)
a10/a30 =1/3
a+9d/a+29d =1/3
3(a+9d)=a+29d
3a+27d=a+29d
3a-a=29d-27d
2a=2d
a=d
sub a=d in (1)
2a+5d=14
2a+5a=14
7a=14
So , a =14/7
a=2
Therefore the first term is 2
 
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S6=42
A10:A30=1:3
n/2(2a+(n-1)d)=42
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3(2a+(n-1)d=42
3(2a+5d)=42
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