Question

If the 3rd and 9th term of an ap are 4&-8 respectively .which term of this ap is 0.​

Answer 1

Answer:

As we know that we are given with the 3rd and 9th term of the AP.

So, let us write an equation for that.

Let the first term of this AP will be equal to a

And the common difference of this AP will be equal to d.

So, as we know that two consecutive terms of any AP differ by a common difference.

So, if the first term of the AP is a. then the second term of this AP will be equal to a + d.

And this goes on and the third term of the AP will be equal to a + 2d. So, by this way the ninth term of the AP will be equal to a + 8d, because each consecutive term increases by d.

But we are given that the 3rd term is 4 and the 9th term is – 8.

So, a + 2d = 4 (1)

And, a + 8d = – 8 (2)

Now to find the value of d. We subtract equation 1 from equation 2. We get,

a + 8d – a – 2d = – 8 – 4

6d = – 12

d = – 2

So, putting the value of d in equation 1. We get,

a + 2(– 2) = 4

a – 4 = 4

a = 8

Now we had to find the term of this AP which is equal to 0.

Let the nth

term of this AP is equal to zero.

As we know that nth

term of any AP is written as a + (n – 1)d. Where a is the first term and d is the common difference.

As we know that the nth

term of this AP is zero.

So, 0 = a + (n – 1)d (3)

Now putting the value of a and d in equation 3. We get,

0 = 8 + (n – 1)(–2)

0 = 8 – 2n + 2

0 = 10 – 2n

2n = 10

So, n = 5

Hence, the 5th term of the given AP is equal to zero.

Step-by-step explanation:

hope it's helpful to u ;)

Answer 2

Answer:

• a
• a n
• a n
• a n =a+(n−1)d
• a n =a+(n−1)da
• a n =a+(n−1)da 3
• a n =a+(n−1)da 3
• a n =a+(n−1)da 3 =4=a+2d ....(1)
• a n =a+(n−1)da 3 =4=a+2d ....(1)a
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−4
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2 a=4−2d
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2 a=4−2d=4+4=8
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2 a=4−2d=4+4=8a
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2 a=4−2d=4+4=8a n
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2 a=4−2d=4+4=8a n
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2 a=4−2d=4+4=8a n =0,a=8,d=−2
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2 a=4−2d=4+4=8a n =0,a=8,d=−20=8+(n−1)×(−2)
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2 a=4−2d=4+4=8a n =0,a=8,d=−20=8+(n−1)×(−2)0=8−2n+2
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2 a=4−2d=4+4=8a n =0,a=8,d=−20=8+(n−1)×(−2)0=8−2n+22n=10
• a n =a+(n−1)da 3 =4=a+2d ....(1)a 9 =−8=a+8d ....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2 a=4−2d=4+4=8a n =0,a=8,d=−20=8+(n−1)×(−2)0=8−2n+22n=10n=5