Question

plzz tell me this yrrrrrrrrr

Answer 1

Heya !!

A.P. :- 3, 15, 27, 39,......

a = 3

d = 15 - 3 = 12

an = a + (n-1) d

a54 = 3 + (54-1) 12

=> 3 + (53) 12

=> 3 + 636

=> 639

Let an be 132 more than it's 54th term

an = a54 + 132

=> 639 + 132

=> 771

=> a + (n-1) d = 771

=> 3 + (n-1) 12 = 771

=> 12 (n-1) = 771 - 3

=> 12 (n-1) = 768

=> n-1 = 768/12

=> n-1 = 64

=> n = 64+1

=> n = 65

Thus, 65th term is 132 more than its 54th term.

Answer 2

Solution:

_____________________________________________________________

Given:

AP: 3 ,15 , 27 , 39....

a = 3,

d = 15 - 3 = 12

n = 54
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To find:

[tex]a_{n} = ? a_{54} + 132 = a_{n} ,[/tex]
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We know that,

$a_n = a+ (n-1)d$

=>$a_{54} = 3 + (54-1)(12)$

=> $a_{54} = 3 +(53)(12)$

=> $a_{54} = 3 + 636$

=> $a_{54} = 639$

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$a_{54} + 132$

=> $639 + 132$

=> $771$

$a_n = 771$

=> $771 = a+ (n-1)d$

=> $771 = 3 + (n-1)(12)$

=> $768 = (n-1)12$

=> $\frac{768}{12} = n-1$

=> $64 = n-1$

=>  ∴ n = 65

The term which is 132 more than the AP's 54th term is its 65th term ,.

($a_{65} = 771$)
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                                     Hope it Helps !!