Question

$\Large\underline\mathbb\red{QUESTION}$


In the following APs, find the missing terms in the boxes :


$(i) \: \tt{2, \boxed \:, \: 26 } \\( ii) \: \: \tt{ \boxed \: ,13, \: \boxed \: ,3} \\( iii) \tt{\: 5, \boxed \: , \boxed \:, \: 9 \frac{1}{2} } \\( iv) \tt{ - 4, \boxed \:, \boxed \: , \boxed \: , \boxed \:, \: 6 } \\ (v) \: \tt{ \boxed \:,38, \boxed \: , \boxed \: , \boxed \: , \: - 22 }$

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Answer 1

★We know that,

In AP the difference between the terms is same.

★ Solution :

$(i) \: \tt{2, \boxed \:, \: 26 } \\ \\ \\ 26 - 2 = 24 \\ \\ \\ \frac{\cancel{24}}{\cancel{2}} = 12 \\ \\ \\ common \: difference = 12 \\ \\ \\ required \:term = 2+12 = \bold{24} \\ \\ \\</p><p>{\pink{\boxed{= 2,14,26}}}\\ \\ \\ ( ii) \: \: \tt{ \boxed \: ,13, \: \boxed \: ,3} \\ \\ \\ There \:is \:one \:box\:between\:13 \: and\:3 \\ \\ \\ 13 - 3 = 10 , common \: difference = \frac{\cancel{10}}{\cancel{2}} = 5 \\ \\ \\ 13+5 = 18 , 13-5 = 8 \\ \\ \\ {\pink{\boxed{= \tt{ 18 \: ,13, \: 8 \: ,3}}}}$

$iii) \tt{\: 5, \boxed \: , \boxed \:, \: 9 \frac{1}{2} } \\ \\ \\ There \:are \:two \:boxes\:in\:between \\ \\ \\ 9 \frac{1}{2} - 5 = 4\frac{1}{2} = \frac{9}{2} \\ \\ \\ common \: difference = \frac{\cancel{9}}{2} \times \frac{1}{\cancel{3}} = \frac{3}{2} \\ \\ \\ 5 + \frac{3}{2} = \frac{10+3}{2} = \frac{13}{2} = 6\frac{1}{2} \\ \\ \\ 6\frac{1}{2} + \frac{3}{2} = \frac{13+3}{2} = \frac{\cancel{16}}{\cancel{2}} = 8 \\ \\ \\ {\pink{\boxed{\bold= { \tt{\: 5, 6\frac{1}{2} \: , 8 \:, \: 9 \frac{1}{2} } }}}}$

$( iv) \tt{ - 4, \boxed \:, \boxed \: , \boxed \: , \boxed \:, \: 6 } \\ \\ \\ 6-(-4) = 6+4 = 10 \\ \\ \\ common\: difference = \frac{\cancel{10}}{\cancel{5}} = 2 \\ \\ \\ -4 +2 = -2 , -2+2 = 0 , 0+2 = 2, 2+2 = 4 \\ \\ \\ {\pink{\boxed{\tt{ - 4, -2\:, 0 \: , 2 \: , 4 \:, \: 6 }}}}$

$(v) \: \tt{ \boxed \:,38, \boxed \: , \boxed \: , \boxed \: , \: - 22 } \\ \\ \\ 38-(-22) = 38+22 = 60 \\ \\ \\ there \:are \: 3 \:boxes\:in \: between \\ \\ \\ common \: difference = \frac{\cancel{60}}{\cancel{4}} = 15 \\ \\ \\ 15 \:is \:subtracted \:from \:each \:term \\ \\ \\ 38 - 15 = 23 , 23 - 15 = 8 , 8-15 = -7\\ \\ \\ first \:term = x , x -15 = 38 \\ x = 38+15 \\ x = 53 \\ \\ \\ {\pink{\boxed{\tt{ 53 \:,38, 23\: , 8 \: , -7 \: , \: - 22 }}}}$