Question

4th term of an arithmetic sequence is 14 and its 9th term is 29 what is the common difference and what is its first term​

Answer 1

Topic

Arithmetic Progression

Given

$4^{th}$term of an Arithmetic Progression is 14 and $9^{th}$ term is 29.

To Find

• The common difference and

• First Term of AP.

Concept Used

$n^{th}$term of an AP is given by :-

$a_n = a + ( n - 1 )d$

where

• a = First term of AP

• n = Number of term

• d = Common Difference

Solution

It is given that $4^{th}$term of AP is 14.

$a_{4}=a+(4-1)d$

$14=a+3d$

Now,

It is given that $9^{th}$term of AP is 29.

$a_{9}=a+(9-1)d$

$29=a+8d$

We need to solve these two obtained equations.

Subtracting equation (1) from equation (2),

$(a+8d)-(a+3d)=29-14$

$a+8d-a-3d=29-14$

$a-a+8d-3d=29-14$

$5d=15$

$d=\dfrac{15}{5}$

$d=3$

Now, put value of 'd' in any obtained equation.

a + 3d = 14

a + 3(3) = 14

a + 9 = 14

a = 14 - 9

a = 5

So, a = 5 and d = 3.

Answer

The first term of AP is 5 and common difference of AP is 3.

Answer 2

Step-by-step explanation:

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given,

4th term of an A.P = 14

9th term of an A.P = 29

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nth term of an A.P = a+(n-1)d

.: 14 = a+(4-1)d

14 = a+3d .... ( 1 )

.: 29 = a+(9-1)d

29 = a+8d ..... (2)

.: (2)-(1)

29-14 = a+8d-a-3d

15 = 5d

d = 15/5

.: d = 3

substituting value of 'd' in ...(1)

14 = a+3(3)

14 = a+9

14-9 = a

.: a = 5

.: first term of A.P = 5 and common difference = 3