Question

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

Answer 1

⠀⠀ıllıllı uoᴉʇnloS ıllıllı

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Given:

• A.P. is 3, 15, 27, 39, …

First term, a = 3

Common difference, d = a2 - a1 = 15 - 3 = 12

We know that,

• an = a + (n - 1) d

Therefore,

a54 = a + (54 - 1) d

⇒ 3 + (53) (12)

⇒ 3 + 636 = 639

a54 = 639

We have to find the term of this A.P. which is 132 more than a54, i.e. 771.

Let nth term be 771.

➠ an = a + (n - 1) d

➠ 771 = 3 + (n - 1) 12

➠ 768 = (n - 1) 12

➠ (n - 1) = 64

➠ n = 65

Therefore, 65th term was 132 more than 54th term.

Alternate Method:

Let nth term be 132 more than 54th term.

n = 54 + 132/12

= 54 + 11

= 65th term

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Answer 2

AnswEr:

$\bold{\underline{\underline{\sf{Given\:AP - }}}}$

$\sf{3,\:15,\:27,\:39...}$

$:\implies\sf\: First\: Term\;(a) = 3$

$:\implies\sf\red{Common\: Difference\:=\: a_{2}\:-\:a_{1}}$

$:\implies\sf\: a_{2} = 15 \:\&\: a_{1}= 3$

$:\implies\sf\: 15 - 3$

$:\implies\small\boxed{\sf{\red{d\:=\:12}}}$

$\rule{120}2$

$\bold{\underline{\sf{By\:Using\: Formula}}}$

$\dag\:\:\small{\underline{\boxed{\sf{\purple{a_{n}= a + (n -1)d}}}}}$

$\sf{We \:Have}\begin{cases}\sf{First\:Term\:(a)\:=\: 3}\\ \sf{Common\: Difference\:(d) \:=12}\\ \sf{n\:=\:54}\end{cases}$

$\:\:\:\:\:\:\;\:\;\bold{\underline{\sf{Putting\:Values:-}}}$

$:\implies\sf\: a_{54} = a + (n -1)d$

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

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

$:\implies\sf\: 3 + 636$

$:\implies\small\boxed{\sf{\purple{a_{54}\:=\: 639}}}$

$\bold{\underline{\sf{54th\:Term\:of\:the\:AP\:is\:639.}}}$

$\rule{120}2$

$:\implies\sf\: 132 + 639$

$:\implies\sf\bold{771}$

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

$:\implies\sf\: 771 = 3 + (n - 1)12$

$:\implies\sf\: 771 - 3 = (n - 1)12$

$:\implies\sf\: 768 = (n -1)12$

$:\implies\sf\: \cancel\dfrac{768}{12} =( n -1)$

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

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

$:\implies\sf\: n = 64 + 1$

$:\implies\small\boxed{\sf{\purple{n\:=\:65}}}$

$\bold{\underline{\sf{\therefore\: 65th\:Term\:is \:132\;more \: than \: of \:54th\;Term. }}}$