Question

If the 3rd and 7th terms of an A.P. are 18 and 30 respectively find the series.
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Answer 1

Answer :

12 , 15 , 18 , 21 , ....

Step-by-step explanation :

$\underline{\bf Arithmetic \ Progression:}$

• It is the sequence of numbers such that the difference between any two successive numbers is constant.

• General form of AP,

     a , a+d , a+2d , a+3d , ..........

• nth term of AP,

       $\boxed{\bf a_n=a+(n-1)d}$

• Sum of n terms,

       $\boxed{\bf S_n=\frac{n}{2}[2a+(n-1)d]} \\\\ \boxed{\bf S_n=\frac{n}{2}[a+l]}$

where

     a - first term

     d - common difference

     n - number of terms

      l - last term

    aₙ - nth term

    Sₙ - sum of n terms

______________________________

Given,

• 3rd term = 18
• 7th term = 30

3rd term :

  a₃ = 18

a + (3 - 1)d = 18

a + 2d = 18   --- (1)

7th term :

   a₇ = 30

a + (7 - 1)d = 30

a + 6d = 30   --- (2)

Subtract (1) from (2),

 a + 6d - (a + 2d) = 30 - 18

 a + 6d - a - 2d = 12

  4d = 12

   d = 12/4

    d = 3

∴ common difference = 3

Substitute d = 3 in equation (1)

   a + 2d = 18

   a + 2(3) = 18

   a + 6 = 18

   a = 18 - 6

   a = 12

first term,

a₁ = 12

Second term,

a₂ = a₁ + d

a₂ = 12 + 3

a₂ = 15

Third term,

a₃ = 18

Fourth term,

a₄ = a₃ + d

a₄ = 18 + 3

a₄ = 21

The series : 12 , 15 , 18 , 21 , ....