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,

S_{6} = 42

=> 6/2 (2a + 5d) = 42

=> 3 (2a + 5d) = 42

=> 2a + 5d = 14 ----- (i)

Again, T_{10}/T_{30} = 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

**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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_{6}= 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)

a

_{10}/a

_{30 }=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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