Question

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•Which term of the A.P. 6, 13, 20, 27 . . . . is 98 more than its 24th term?
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Answer 1

✠ Correct Question :-

• Which term of the A.P. 6 , 13 , 20 , 27 . . . . is 97

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✪ Given :-

• First term = 6
• Last term = 97
• Common difference = 7

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✪ To Find :-

• Which term of the A.P. 6 , 13 , 20 , 27 . . . . is 97

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✪ Solution :-

By using A.P. formula

$\large \red \bigstar \: \boxed{ \green{\bf A_n = a + (n - 1)d}}$

Here

• a = 6
• d = 7
• Aₙ = 97
• n = ?

Substitute values in formula

$\longmapsto \sf97= 6 + (n - 1)7$

$\longmapsto \sf97 - 6 = (n - 1)7$

$\longmapsto \sf91 = (n - 1)7$

$\longmapsto \sf n - 1 = \frac{91}{7}$

$\longmapsto \sf n - 1 = 13$

$\longmapsto \sf n = 14$

∴ 14th term of the A.P. is 97

Answer 2

Given :-

First term = 6

Last term = 97

Common difference = 7

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To Find :-

Which term of the A.P. 6 , 13 , 20 , 27 . . . . is 97

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Solution :-

By using A.P. formula

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

Here

a = 6

d = 7

Aₙ = 97

n = ?

Substitute values in formula

⟹97=6+(n−1)7

⟹97−6=(n−1)7

⟹91=(n−1)7

⟹n−1= $\frac{7}{91}$

⟹n−1=13

⟹n=14

∴ 14th term of the A.P. is 97

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