Question

which of the following sequences is an AP?
if they are AP , find the common difference.
$1) 2, \frac{5}{2} , 3, \frac{7}{2} ..$
2) -10 ,-6,-2 ,2​

Answer 1

Answer

$Here , \: the \: 1 st \: term \: \\ \\ t 1 = 2$

$t 2 - t 1 = \frac{5}{2} - 2 \\ \\ = \frac{5 - 4}{2} = \frac{1}{2}$

$t \: 3 - t \: 2 \: = 3 - \frac{5}{2} \\ = \frac{6 - 5}{2} = \frac{1}{2}$

$t \: 4 \: - t \: 3 \: \frac{7}{2} - 3 \\ \\ = \frac{7 - 6}{2} = \frac{1}{2}$

The given sequences is AP because

• The common difference between any two consecutive Term is constant .

The common difference

$d \: = \frac{1}{2}$

2 nd question answer

Here ,

$the \: 1st \: term \: \\ \\ = t \: 1 = - 10$

$t \: 2 - \: t \: 1 \:$

= -6 -(-10) = -6 +10 = 4

=t 3 - t 2 = -2 - (-6)-2 +6

= 4

=t 4 - t 3 = 2 -(-2)=2+2= 4

This sequences is also an AP because

• The common difference between any two consecutive Term is constant .

The common difference

d = 4

Answer 2

$Here,the1sttermt1=2</p><p>\begin{gathered}t 2 - t 1 = \frac{5}{2} - 2 \\ \\ = \frac{5 - 4}{2} = \frac{1}{2} \end{gathered}t2−t1=25−2=25−4=21</p><p>\begin{gathered}t \: 3 - t \: 2 \: = 3 - \frac{5}{2} \\ = \frac{6 - 5}{2} = \frac{1}{2} \end{gathered}t3−t2=3−25=26−5=21</p><p>\begin{gathered}t \: 4 \: - t \: 3 \: \frac{7}{2} - 3 \\ \\ = \frac{7 - 6}{2} = \frac{1}{2} \end{gathered}t4−t327−3=27−6=21</p><p>$

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