Question

If a = 10 and d=10, then first four
terms will be​

Answer 1

Answer:

The first four terms of the AP would be 10,20,30 and 40

Step-by-step explanation:

Given:

• First term(a) = 10
• Common difference (d) = 10
• n = 4

To Find:

• First four terms

Solution:

As we know, the first term would be 10, then

• a(n) = a + (n-1)d

• a(2) = 10 + (2-1)(10)

• a(2) = 10 + 10

• a(2) = 20

The third term would be,

• a(n) = a + (n-1)d

• a(3) = 10 + (3-1)(10)

• a(3) = 10 + 20

• a(3) = 30

And the fourth term would be,

• a(n) = a + (n-1)d

• a(4) = 10 + (4-1)(10)

• a(4) = 10 + 30

• a(4) = 40

∴ The first four terms of the ap would be 10, 20,30 and 40.

____________________________

Answer 2

Given that , First term of an A.P is ( a ) = 10 & Common Difference is ( d ) = 10 .

Need To Find : The First four terms of an A.P ( or Airthmetic Progression ).

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀¤ Finding First four term of an A.P :

As , We know that ,

⠀⠀⠀⠀⠀Formula for an AIRTHMETIC PROGRESSION :

$\qquad \star \:\:\:\underline {\boxed {{\pink{\frak { a_n \:=\: a \: + \: \bigg( \:n - 1\: \bigg) \: d \:}}}}}$

⠀⠀⠀⠀⠀Here a is the first term of an A.P , n is the nth term of an A.P & d is common Difference.

⠀⠀We already know that , First Term of an A.P ( or Airthmetic Progression ) is 10 .

⠀⠀⠀➠ Second term of an A.P :

$\qquad \dashrightarrow \sf a_n \:=\: a \: + \: \bigg( \:n - 1\: \bigg) \: d$

$\qquad \dashrightarrow \sf a_2 \:=\: 10 \: + \: \bigg( \:2 - 1\: \bigg) \: 10$

$\qquad \dashrightarrow \sf a_2 \:=\: 10 \: + \: \bigg( \:1\: \bigg) \: 10$

$\qquad \dashrightarrow \sf a_2 \:=\: 10 \: + \: \: 10$

$\qquad \dashrightarrow \sf a_2 \:=\: 20$

$\qquad \therefore \:\: \underline{\boxed{\purple {\pmb {\frak{ \:\: a_2\:\:(\:or \:Second \:term\:)\:=\: 20 \:\:}}}}}$

• Second term of an A.P is 20 .

⠀⠀⠀➠ Third term of an A.P :

$\qquad \dashrightarrow \sf a_n \:=\: a \: + \: \bigg( \:n - 1\: \bigg) \: d$

$\qquad \dashrightarrow \sf a_3 \:=\: 10 \: + \: \bigg( \:3 - 1\: \bigg) \: 10$

$\qquad \dashrightarrow \sf a_3 \:=\: 10 \: + \: \bigg( \:2\: \bigg) \: 10$

$\qquad \dashrightarrow \sf a_3 \:=\: 10 \: + \: \: 20$

$\qquad \dashrightarrow \sf a_3 \:=\: 30$

$\qquad \therefore \:\: \underline{\boxed{\purple {\pmb {\frak{ \:\: a_3\:\:(\:or \:Third \:term\:)\:=\: 30 \:\:}}}}}$

• Third term of an A.P is 30 .

⠀⠀⠀➠ Fourth term of an A.P :

$\qquad \dashrightarrow \sf a_n \:=\: a \: + \: \bigg( \:n - 1\: \bigg) \: d$

$\qquad \dashrightarrow \sf a_4 \:=\: 10 \: + \: \bigg( \:4 - 1\: \bigg) \: 10$

$\qquad \dashrightarrow \sf a_4 \:=\: 10 \: + \: \bigg( \:3\: \bigg) \: 10$

$\qquad \dashrightarrow \sf a_4 \:=\: 10 \: + \: \: 30$

$\qquad \dashrightarrow \sf a_4 \:=\: 40$

$\qquad \therefore \:\: \underline{\boxed{\purple {\pmb {\frak{ \:\: a_4\:\:(\:or \:Fourth \:term\:)\:=\: 40 \:\:}}}}}$

• Fourth term of an A.P is 40 .

$\qquad \therefore \:\:\underline {\sf \:Hence,\:First \:four \:terms \:of \:an \: A.P \:are \:\:\bf 10 \:,\:20\:,\:30\:,\:and\:40\:.}$