Question

*If 2nd and 7th term of an AP are 12 and -3 respectively, then which term of this AP is zero?*


1️⃣ 1st
2️⃣ 5th
3️⃣ 6th
4️⃣ 10th​

Answer 1

Answer:

6th term is Zero.

Step-by-step explanation:

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Answer 2

Answer

• 6th term = 0. ( option 3 )

Given

• 2nd and 7th term of an AP are 12 and -3.

To Do

• To find which term of the AP will be zero.

Step By Step Explanation

Formula Used :

$\red \bigstar \: \: \: \: \: \: \:\underline{ \boxed{ \bold{a_{n} = a + (n - 1)d}}}$

Substituting the values :

Let's substitute the values !!

For 2nd term ⤵

$\longmapsto\tt a_{2} = a + (2 - 1)d \\ \\\longmapsto \bold{ 12 = a + d \: \: \: \: \: \: - \: \: \: eq.1}$

For 7th term ⤵

$\longmapsto \tt a_{7} = a + (7 - 1)d \\ \\ \longmapsto \bold{ - 3 = a + 6d \: \: \: \: \: \: - \: \: \: eq.2}$

By subtracting :

Let's substrate 1st equation by 2nd.

$\sf12 = a + d \\ \sf - 3 = a + 6d \\ \sf( + ) \: \: \: \: ( - )( - ) \\ - - - - - - \\ \sf{15 = -5d} \implies \: \bold{d = - 3}$

Now, a will be => 12 = a + ( -3 ) => 15.

Now, we need to find the term of the AP which is 0.

So let's do it !!

$\longmapsto \bold{\: a_{n} = a +( n - 1)d} \\ \\\longmapsto \sf 0 = 15 + (n - 1)( - 3) \\ \\\longmapsto \sf - 15 = (n - 1)( - 3) \\ \\\longmapsto \sf \cancel\cfrac{ - 15}{ - 3} = (n - 1) \\ \\\longmapsto \sf 5 = (n - 1) \\ \\\longmapsto \sf 5 + 1 = n \\ \\ \longmapsto { \underline { \boxed{\bold{ \green{6 = n}}}}} \: \: \: \: \: \: \: \: \: \: \: \: \red\bigstar$

Therefore, 6th term = 0. ( option 3 ).

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