Question

the sum of n terms of an AP is 6nsquare+6n, then the fourth term of the series is​

Answer 1

Given

$\sf\longmapsto{S_n = 6n^2 + 6n}$⠀....[1]

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To find

$\sf\longmapsto{Fourth\: term\: of\: the\: series.}$

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Solution

• We can write

$\tt\longmapsto{S_{n-1} = 6(n - 1)^2 + 6(n - 1)}$⠀...[2]

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• We know that

$\tt\longmapsto{S_n - S_{n-1} = nth\: term}$

Therefore,

$\tt\longmapsto{S_4 - S_{4-1} = 4th\: term}$

$\tt\longmapsto{S_4 - S_3 = 4th\: term}$⠀...[3]

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★ From [1], [2] and [3], we get

$\small\tt:\implies\: \: \: \: \: \: \: \: {S_4 - S_3 = (6n^2+6n) - [6(n - 1)^2 + 6(n - 1)]}$

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$\small\tt:\implies\: \: \: \: \: \: \: \: {4th\: term = (6n^2 + 6n) - [6(n^2 + 1 - 2n)+ 6n - 6]}$

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$\small\tt:\implies\: \: \: \: \: \: \: \: {4th\: term = (6n^2 + 6n) - (6n^2 + 6 - 12n + 6n -6)}$

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$\small\tt:\implies\: \: \: \: \: \: \: \: {4th\: term = 6n^2 + 6n - 6n^2 + 6n}$

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$\small\tt:\implies\: \: \: \: \: \: \: \: {4th\: term = 6n + 6n}$

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$\tt:\implies\: \: \: \: \: \: \: \: {4th\: term = 12n}$

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Hence, the fourth term is 12n.

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