Question

ɪғ ᴛʜᴇ 3ʀᴅ ᴀɴᴅ ᴛʜᴇ 9ᴛʜ ᴛᴇʀᴍs ᴏғ ᴀɴ ᴀ.ᴘ. ᴀʀᴇ 4 ᴀɴᴅ − 8 ʀᴇsᴘᴇᴄᴛɪᴠᴇʟʏ, ᴛʜᴇɴ ᴡʜɪᴄʜ ᴛᴇʀᴍ ᴏғ ᴛʜɪs ᴀ.ᴘ ɪs ᴢᴇʀᴏ. to?​

Answer 1

Answer:

2nd term

Step-by-step explanation:

Let the first term of the AP be a and common difference is d.

so the third term=a+2d=4

and 9th term=a+8d=-8

now 9th term-3rd term=-8-4

=> a+8d-a-2d=-12

=> 6d=12

=> d=12/2=6

so a=4-2d=4-12=-6

Thus, a+d=0

=> a+(2-1)d=0.

Therefore the second term of the AP is 0.

(answer)

hope you got it.

Answer 2

$\huge\sf\pink{Answer}$

☞ The 5th term of the AP is 0

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$\huge\sf\blue{Given}$

✭ 3rd term of an AP is 4

✭ 9th term of an AP is -8

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$\huge\sf\gray{To \:Find}$

◈ Which term of the AP is see zero?

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$\huge\sf\purple{Steps}$

We know that,

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

☯ $\underline{\textsf{As Per the Question}}$

➳ $\sf a_3 = a+2d=4 \:\:\: -eq(1)$

➳ $\sf a=4-2d\:\:\:-eq(2)$

➳ $\sf a_9=a+8d=-8\:\:\: -eq(3)$

Substituting eq(2) in eq(3)

➝ $\sf a_9=a+8d=-8$

➝ $\sf 4-2d+8d=-8$

➝ $\sf 6d=-8-4$

➝ $\sf 6d=-12$

➝ $\sf d=\dfrac{-12}{6}$

➝ $\sf \green{d=-2}$

Substituting value of d in eq(2)

➢ $\sf a=4-2d$

➢ $\sf a=4-2(-2)$

➢ $\sf a=4+4$

➢ $\sf \red{a=8}$

Now that we have the value of a & d we shall find the nth term of an ap,for that we shall use,

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

Substituting the values

»» $\sf a_n=a+(n-1)d$

»» $\sf 0=8+(n-1)(-2)$

»» $\sf -8=(n-1)(-2)$

»» $\sf \dfrac{-8}{-2}=n-1$

»» $\sf 4=n-1$

»» $\sf 4+1=n$

»» $\sf \orange{n=5}$

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