Question

which term of the AP:3,15,27,39,...will be 132 more than 54th term.please give step by step​

Answer 1

Answer:

65th term

Step-by-step explanation:

In APs, nth term is given by a + (n - 1)d , a is first term and d is common difference.

Here,

a = first term = 3

d = 15 - 3 = 12

Let nth term be 132 more than 54th term

= > 3 + (n - 1)12 = 132 + {3 + (54 - 1)12}

= > 3 + 12n - 12 = 132 + 3 + (53)12

= > 12n - 9 = 135 + 636

= > 12n = 135 + 636 + 9

= > 12n = 780

= > n = (780/12)

= > n = 65

65th term is 132 more than 54th term.

Answer 2

QUESTION :-

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

NOTATIONS USED :-

$\: \: \: \: \: \: \: \: \: \: \: \: \: { \text{ \pink{a = first \: term}}}$

$\: \: \: \: \: \: \: \: \: { \text { \pink{d \: = common \: difference}}}$

$\: \: \: \: \: \: \: \: \: \: \: { \text{ \pink{n \: = number \: of \: terms}}}$

GIVEN :-

• A. P = 3 , 25 , 37 , 39

Where,

• a = 3
• d = 15 - 3 = 12
• n = 54

SOLUTION :-

${ \boxed{ \boxed{\mathtt{ \blue{ a_{54} = a +(n - 1)d}}}}}$

By putting given values in the above formula

${ \sf{\implies \: 3 + (54 - 1) \times 12}}$

${ \sf{ \implies \: 3 + 53 \times 12}}$

${ \sf{ \implies \: 639}}$

132 is more then it's 54th term so

${ \sf{ \implies - 639 + 132}}$

${ \sf{ \implies \: a_{n} = 771}}$

${ \sf{ \implies{a + (n - 1)d}}}$

${ \sf{ \implies \: 3 +(n - 1)12}}$

${ \sf{ \implies \: 64 = n - 1}}$

${ \boxed{ \bf{ \green{ \implies \: n = 65}}}}$

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

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