Question

Which term of the AP 3,15,27,39,.....will be 132 more than its 54 th term ?​

Answer 1

AnswEr :

⋆ Given AP : 3, 15, 27, 39...

⋆ First Term ( a ) = 3

⋆ Difference ( d ) = 15 - 3 = 27 - 15 = 12

⋆ Which term is 132 more than 54th term.

• we will find the 54th term of AP :

⇒ Nth Term = a + (n - 1)d

⇒ a₅₄ = 3 + (54 - 1) × 12

⇒ a₅₄ = 3 + 53 × 12

⇒ a₅₄ = 3 + 636

⇒ a₅₄ = 639

we have 54th Term as 639, Now we will add 132 to find the New Nth Term ;

⇝ Nth Term = 639 + 132 = 771

• we will find which term is 771 now :

⇒ Nth Term = a + (n - 1)d

⇒ 771 = 3 + (n - 1)12

⇒ 771 - 3 = (n - 1)12

⇒ 768 = (n - 1)12

• Dividing Both term by 12

⇒ 64 = (n - 1)

⇒ n = 64 + 1

⇒ n = 65

∴ 65th term is 132 more than 54th term.

Answer 2

Answer:

$\large\bold\red{65th\:term}$

Step-by-step explanation:

Given,

An A.P such that,

$\bold{3,15,27,39,...........}$

Here,

• First term, a = 3
• Common difference, d = (15-3) = 12

Now,

We know that,

nth term of an AP is given by,

$\large \boxed{ \purple{a_{n} = a + (n - 1)d}}$

Therefore,

We have,

$= > a_{54} = a + (54 - 1)d \\ = > a_{54} = a + 53d \\ = > a_{54} = 3 +( 53 \times 12) \\ = > a_{54} = 3 + 636 \\ = > a_{54} = 639$

Now,

Let, 132 more than 54th term be x

Therefore,

$= > x = 639 + 132 \\ = > x = 771$

Now,

We have to find that which term 'x' is .

But,

We know that,

From formula of general term of that AP,

$= > x = 3 + 12(n -1 )$

Therefore,

We have,

$= > 3 + 12(n - 1) = 771 \\ \\ = > 12(n - 1) = 771 - 3 \\ \\ = > 12(n - 1) = 768 \\ \\ = > n - 1 = \cancel{ \frac{768}{12} } \\ \\ = > n - 1 = 64 \\ \\ = > n = 64 + 1 \\ \\ = > n = 65$

Hence,

The required number is 65th term of the A.P