Question

Consider an arithmetic sequence whose 6th term is 40 and 9th term is 58.
A.find its common difference?
B.find out its first term?

Answer 1

given,

a + 5d = 40---------eq.1

and

a + 8d = 58---------eq.2

solution-----

a = 40 - 5d---------- eq.3

putting eq.3 in eq.2 ---

40- 5d + 8d = 58

3d = 18

d = 18/3

d = 6 (common difference of A.P)

now put the value of d in any of the equations i .e. eq.1 and 2----

a + 5×6 = 40

a = 40 - 30

a = 10

A. common difference is 6.

B. first term is 10.

I hope it will help you❤

Answer 2

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♠ Given :-

For An Arthematic Progression (A.P.) :

• 6th Term =  $\sf a_6$ = 40

• 9th Term =  $\sf a_9$ = 58

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♠ To Find :-

• Common Difference

• First Term

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♠ Main Formula :-

Formula For nth Term of an A.P :

 $\Large\bf a_n =a+(n-1)d$

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♠ Solution :-

We Have,

$\large \sf a_6 = 40 \\ \\ :\longmapsto \large\bf a + 5d = 40 \: \:- - - (1)$

$\large \sf a_9 = 58 \\ \\ \bf: \large\longmapsto a + 8d = 58 \: \: - - - (2)$

$\Large \bigstar \: \underline{ \pmb{ \mathfrak{ \text{S}ubtracting \: (1) \: from \: (2) }}}$

$:\longmapsto \sf3d = 18 \\ \\ \Large \purple{ :\longmapsto  \underline {\boxed{{\bf d = 6} }}}$

$\Large \bigstar \: \underline{ \pmb{ \mathfrak{ Putting \: \text{V}alue \: of \: \text{d}\: in \: (1)}}}$

$:\longmapsto \sf a + 5 \times 6 = 40 \\ \\ :\longmapsto \sf a + 30 = 40 \\ \\ :\longmapsto \sf a = 40 - 30 \\ \\ \Large\purple{ :\longmapsto  \underline {\boxed{{\bf a = 10} }}}$

Hence ,

• Common Difference = 6

• First Term = 10

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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