Question

Which term of the AP :21,18,15,......is -81?, is any term 0​

Answer 1

Answer:

In an AP

a=21

d=-3

l=-81

a+(n-1)d=-81

21+(n-1)(-3)=-81

(n-1)(-3)=-81-21

(n-1)(-3)=-102

(n-1)=-102/-3

(n-1)=34

n=34+1

n=35

Step-by-step explanation:

also a+(n-1)d=0

21+(n-1)(-3)=0

(n-1)(-3)=-21

n-1=-21/-3

n-1=7

n=7+1

n=8

therefore 8th term is zero

Answer 2

Answer:

$\huge{\pink{\boxed{\green{\sf{n_{1}=35}}}}}$

$\huge{\pink{\boxed{\green{\sf{n_{2}=8}}}}}$

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

$\: \: \: \: \: \: \: \: { \orange{given}} \\ { \pink{ \boxed{ \green{a.p = 21.18.15.......}}}} \\ { \pink{ \boxed{ \green{a_{1} = 21 }}}} \\ { \pink{ \boxed{ \green{d= - 3 }}}} \\ { \pink{ \boxed{ \green{a_{n} = - 81 }}}} \\ \\ { \blue{to \: find}} \\ { \purple{ \boxed{ \red{n= }}}}$

☆ According to given question:

$\to a _{n} = a + (n - 1)d \\ \to - 81 =21 + (n_{1} - 1) \times - 3 \\ \to - 81 - 21 = - 3n_{1} + 3 \\ \to - 102 = - 3n_{1}+ 3 \\ \to - 102 - 3 = - 3n_{1} \\ \to - 105 = - 3n_{1} \\ \to n_{1} = \frac{ - 105}{ - 3} \\ {\pink{ \boxed{ \green{ \therefore n_{1} = 35 }}}}$

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Again,

☆ By taking an = 0

$\to a_{n} = a + ( n_{2} - 1)d \\ \to 0 = 21 + ( n_{2} - 1) \times - 3 \\ \to - 21 = ( n_{2} - 1) \times - 3 \\ \to \frac{ - 21}{ - 3} = n_{2} - 1 \\ \to 7 + 1 = n_{2} \\ { \pink{ \boxed{ \green{ \therefore n_{2} = 8 }}}}$

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