Question

In an AP the the third and seventh terms are 14 and -2 respectively which term of the AP is -14​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:n=10

Step-by-step explanation:

Given: third term,a3=14

and seventh term,a7=-2

also given that,an=-14

We know that,an=a+(n-1)d

therefore,

a3=a+(3-1)d

14=a+2d............(1)

and

a7=a+(7-1)d

-2=a+6d............(2)

On subtracting equation (1) from (2)...

We get,

= -2-14=a+6d-a+2d

= -16=4d

= d=-16/4

= d= -4

From the equation (1),we can write,

14=a+2d

14=a+2×(-4)

14=a+ (-8). { + × - =- }

a=14+8

a= 22

Let the nth term of this AP be -14

an=a+(n-1)d

-14=22+(n-1)(-4)

-14=22-4n+4

-14=26-4n

-4n=-14-26

= -40

n= -40/-4

n=10

Therefore,10th term of this arithmetic progression is -14

•: a10 = -14