Question

which term of the AP 3 15 2739 will be 132 more than its 54th term​

Answer 1

Answer:

Thanks for the question!

Answer 2

Given

• A.P. is 3, 15, 27, 39, ...

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To find

• The term of the given A.P. that is 132 more than its 54th term.

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Solution

• Before solving this question, let us learn more about the question.

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★ We know that

$\rightarrow{\boxed{\sf{\orange{a_n = a + (n - 1)d}}}}$⠀⠀....[1]

$\rightarrow{\boxed{\sf{\orange{a_{54} = a + 53d}}}}$⠀⠀.....[2]

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→ Given A.P. is 3, 15, 27, 39, ...

Clearly, it's

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

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$\tt\longrightarrow{}$ Let $\sf\pink{n^{th}}$ term of the A.P. be 132 more than its 54th term.

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★ According to the question

$\tt\longmapsto{a_n = 132 + a_{54}}$

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$\small\tt\longmapsto{a + (n - 1)d = 132 + (a + 53d)}\: \: \: \: \: \: \: {\bf{\bigg\lgroup{From [1] and [2]}{\bigg\rgroup}}}$

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$\tt\longmapsto{3 + (n - 1)12 = 132 + [3 + 53(12)]}$

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$\tt\longmapsto{3 + 12n - 12 = 132 + [3 + 636]}$

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$\tt\longmapsto{12n - 9 = 132 + 639}$

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$\tt\longmapsto{12n - 9 = 771}$

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$\tt\longmapsto{12n = 771 + 9}$

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$\tt\longmapsto{12n = 780}$

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$\tt\longmapsto{n = \cancel{\dfrac{780}{12}}}$

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$\bf\longmapsto{n = 65}$

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Hence, 65th term of the given A.P. is 132 more than its 54th term.

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