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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- -6
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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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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????? =6/2(2a+(6 - 1)d)
Or,????? 6/2(2a+5d)=42
Or,??? 6(a+5d)=84
Or,?????? 2a+5d=42/6
Or,??? 2a+5d=14????? ......(1)
By question, a+9d/a+29d ,3a-a=29d-27d,2a =2d
2a-2d=0 ...........(2), Equation (1)-(2)
We get, 7d=14, d=2,Put the value of d in eq- (2)
2a-2d=0 ,2a-2?2=0 ,2a-4=0, 2a=4, a=4/2, a=2
AP=2,4,6,8,10.........
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