Question

If tn=0 for an arithmetic sequence 84, 78, 72.... then the value of n is

Answer 1

Given : Arithmetic Progression is 84, 78, 72 . . .

To find : Value of n at which the term of AP is 0

Solution :

Arithmetic progression is a sequence of numbers in which common difference between two consecutive terms is constant through out the sequence.

We have first term of AP = 84

and it's common difference = a2 - a1 = 78 - 84 = -6

We have to find value of n at which an is 0.

We have a general representation of term of AP as:

• an = a + ( n - 1 ) d

Here,

• an = nth term
• a = First term
• d = Common difference
• n = Number of terms

By substituting the known values in the general equation, we get :

→ 0 = 84 + ( n - 1 ) ( - 6 )

→ -84 = ( n - 1 ) ( - 6 )

→ -84 / -6 = ( n - 1 )

→ 1 4 = n - 1

→ 14 + 1 = n

→ 15 = n

So the 15th term of given AP is 0.

Required answer is 15.

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Answer 2

SOLUTION

GIVEN

$\sf{t_n = 0 }$

For the arithmetic sequence 84, 78, 72....

TO DETERMINE

The value of n

EVALUATION

The given arithmetic sequence 84, 78, 72....

First term = a = 84

Common Difference = d = 78 - 84 = - 6

The n th term of the sequence is

$\sf{ =t_n }$

$\sf{ =a + (n - 1)d}$

$\sf{ =84 - 6 (n - 1)}$

$\sf{ =84 - 6 n + 6}$

$\sf{ =90 - 6 n }$

By the given condition

$\sf{90 - 6 n = 0}$

$\sf{ \implies \: 6 n = 90}$

$\sf{ \implies \: n = 15}$

FINAL ANSWER

Hence the required value of n = 15

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