Question

The first term of an a.p. is 9 and fifth term is 14. write common difference of the ap

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Common\:Difference=\frac{5}{4}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies First \: term( a_{1}) = 9 \\ \\ \tt: \implies Fifth \: term( a_{5}) = 14 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Common \: Difference = ?$

• According to given question :

$\tt \circ \: a_{1} =9 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies a_{5} = 14 \\ \\ \tt: \implies a_{1} + 4d = 14 \\ \\ \text{Putting \: value \: of \: a}_{1} \\ \tt: \implies 9 + 4d = 14 \\ \\ \tt: \implies 4d = 14 - 9 \\ \\ \tt: \implies 4d = 5 \\ \\ \green{\tt: \implies d = \frac{5}{4}} \\ \\ \blue{ \bold{Some \: related \: formula : }} \\ \orange{ \tt \circ \: a_{n} = a +( n - 1)d} \\ \\ \orange{ \tt \circ \: s_{n} = \frac{n}{2}(2a + (n - 1)d)}$

Answer 2

Answer:

d = 5/4

Note:

★ AP ( Arithmetic Progression ) : A sequence in which the difference between the consecutive terms are same .

★ The number term of an AP is given by ;

T(n) = a + (n-1)d , where a is the first term and d is the common difference of the AP .

Solution:

Given : First term , a = 9

Fifth term , T(5) = 14

To find : Common difference , d = ?

Here,

The fifth term of the AP is 14 .

Thus,

If T(n) = a + (n - 1)d

=> T(5) = a + (5 - 1)d

=> T(5) = a + 4d

=> 14 = 9 + 4d { a = 9 , T(5) = 14 }

=> 4d = 14 - 9

=> 4d = 5

=> d = 5/4

Hence,

The common difference (d) of the AP is 5/4 .