Question

Find the five term:– a=13 , d=–12

Answer 1

AnsWer :

→ 13 , 1 , - 11 , - 23 , - 35

SolutioN :

We have ,

• a = 13.
• d = - 12.

Let,

• " a " first term.
• " d " common difference.

We know, General Formula of AP

→ a , a + d , a + 2d , a + 3d .....

Now,Putting given.

→ 13 , 13 + ( -12 ) , 13 + 2( - 12 ) , 13 + 3( -12 ) + 13 + 4( -12 )

→ 13 , 1 , - 11 , - 23 , - 35 .

★ Our AP become ( 13 , 1 , - 11 , - 23 , - 35 )

Verification :

• a1 = 13.
• a2 = 1.
• a3 = - 11.
• a4 = - 23.
• a5 = - 35.
• a6 = - 47.

† an = a + ( n - 1 )d.

So,

• a1 = a → 13.
• a2 = a + d → 13 - 12 = 1.
• a3 = a + 2d → 13 + 2( - 12 ) = - 11.
• a4 = a + 3d → 13 + 3( -12 ) = - 23.
• a5 = a + 4d → 13 + 4( - 12 ) = - 35.

Answer 2

Question :–

• Find first five term of A.P. , if a = 13 and d = -12

ANSWER :–

• First five term are 13 , 1 , -11 , -23 , -35

EXPLANATION :–

GIVEN :–

• First term of A.P. (a) = 13

• Common difference of A.P. (d) = -12

TO FIND :–

• First five term of A.P.

SOLUTION :–

• We know that nth term of A.P. is –

$\\ \implies \large{ \red{ \boxed{ \bold{ T_{n} = a + (n - 1)d }}}}$

• So that –

$\\ \: \: \longrightarrow \: \: { \bold{First \: \: term( T_{1}) = a + (1 - 1)d }}$

$\\ \: \: \: \: \: \: \: \: \: { \huge{.}} \: \: { \bold{First \: \: term( T_{1}) = a = 13 }}$

$\\ \: \: \longrightarrow \: \: { \bold{Second \: \: term( T_{2}) = a + (2 - 1)d }}$

$\\ \: \: \: \: \: \: \: \: \: { \huge{.}} \: \: { \bold{Second \: \: term( T_{2}) = a + d }}$

$\\ \: \: \: \: \: \: \: \: \: { \huge{.}} \: \: { \bold{Second \: \: term( T_{2}) = 13 - 12 = 1 }}$

$\\ \: \longrightarrow \: \: { \bold{Third \: \: term( T_{3}) = a + (3 - 1)d }}$

$\\ \: \: \: \: \: \: \: \: \: { \huge{.}} \: \: { \bold{Third \: \: term( T_{3}) = a + 2d = 13 +2 ( - 12) }}$

$\\ \: \: \: \: \: \: \: \: \: { \huge{.}} \: \: { \bold{Third \: \: term( T_{3}) = - 11 }}$

$\\ \: \longrightarrow \: \: { \bold{Fourth \: \: term( T_{4}) = a + (4 - 1)d }}$

$\\ \: \: \: \: \: \: \: \: \: { \huge{.}} \: \: { \bold{Fourth \: \: term( T_{4}) = a + 3d = 13 +3 ( - 12) }}$

$\\ \: \: \: \: \: \: \: \: \: { \huge{.}} \: \: { \bold{Fourth \: \: term( T_{4}) = - 23 }}$

$\\ \: \longrightarrow \: \: { \bold{Fifth \: \: term( T_{5}) = a + (5 - 1)d }}$

$\\ \: \: \: \: \: \: \: \: \: { \huge{.}} \: \: { \bold{Fifth \: \: term( T_{4}) = a + 4d = 13 +4 ( - 12) }}$

$\\ \: \: \: \: \: \: \: \: \: { \huge{.}} \: \: { \bold{Fifth \: \: term( T_{5}) = - 35 }}$