Question

The sum of first 6 term of an AP is 42 the ratio of its 10th term to 28 term is 1:3 calculate the first and 13th term of the AP

Answer 1

Let the first term of the A.P be a .

Let the common difference of the A.P be d .

Sum of n terms is given by :

Sn = n / 2 [ 2 a + ( n - 1 ) d ]

So Sum of 6 terms here would be :

S = 6 / 2 [ 2 a + ( n - 1 ) d ]

⇒ S = 3 [ 2 a + ( n - 1 ) d ]

⇒ 3 [ 2 a + ( n - 1 ) d ] = 42

⇒ 2 a + ( n - 1 ) d = 42 / 3

⇒ 2 a + ( n - 1 ) d = 14

n = 6 .

So 2 a + ( 6 - 1 ) d = 14

⇒ 2 a + 5 d = 14 ....................( 1 )

Now ratio of 10 th term to 28 th term = 1 : 3

10 th term = a + ( 10 - 1 ) d = a + 9 d

28 th term = a + ( 28 - 1 ) d = a + 27 d

Ratio = 1 : 3

So we can write :

$\mathsf{\frac{a+9d}{a+27d}=\frac{1}{3}}\\\\\mathsf{\implies 3(a+9d)=a+27d}\\\\\mathsf{\implies 3a+27d=a+27d}\\\\\mathsf{\implies 3a=a}\\\\\mathsf{\implies 2a=0}$

a = 0

Putting this in ( 1 ) :

2 ( 0 ) + 5 d = 14

⇒ 5 d = 14

⇒ d = 14/5

First term = 0

13 th term = a + ( 13 - 1 ) d

                ⇒ 0 + 12 × d

                ⇒ 0 + 12 × 14/5

                ⇒ 168/5

                ⇒ 33.6

The 13 th term is 33.6 .

The first term is 0.

Answer 2

Let the first term be a and common difference be d.

∴   S6 = 42

⇒ n/2 [ 2a + ( n - 1 ) d ] = 42

⇒ 6/2 [ 2a + ( 6 - 1 ) d ] = 42

⇒ 3 [ 2a + 5d ] = 42

⇒ 2a + 5d = 14........(i)

NOW,

a10 / a28 = 1 / 3

⇒ a + 9d / a + 27d = 1 / 3

⇒ 3 ( a + 9d ) = a + 27d

⇒ 3a + 27d = a + 27 d

⇒ 3a - a = 0

⇒ 2a = 0

⇒a = 0

       ∴ First term = 0

Putting a = 0 in (i) ⇒ 0 + 5d = 14

⇒ 5d = 14

⇒ d = 14/5

NOW,  Thirteenth term = a + 12d = 0 + ( 12 × 14/5 )

                                                       = 12 × 14 / 5

                                                       = 168 / 5

                                                        = 33.6

ANS:-      FIRST TERM = 0

               THIRTEENTH  TERM = 33.6