Question

13, 23, 33,... is an arithmetic sequence a) find common difference
b) find the first three digit number of this sequence ​

Answer 1

AnswEr:-

Your Answer is,

• a) 10.

• b) 103.

ExplanaTion:-

Given:-

• An Arithmetic Sequence 13, 23, 33,......

To Find:-

• a) The Common Difference.

• b) The first three digit number of the sequence.

Formula Used:-

$\pink {\longrightarrow \boxed{ \red {\bf d = a_2 - a.}}}$

$\pink{\longrightarrow\boxed{ \red{ \bf a_n = a + (n - 1)d.}}}$

Where,

• d = Common Difference.

• $\tt a_2$ And a Are first and second term respectively.

• $\tt a_n = n^{th}$ Term.

• n = Number of terms.

So Here,

• d = ?? [To Find].

• $\tt a_2$ and a are 13 and 23 respectively.

• $\tt a_n$ is a three digit number i.e. greater than 99.

• n = ?? [To Find].

Now,

In (a) we have to find common difference (d) so by above formula we get,

$\tt\mapsto d = a_2 - a.$

$\tt\mapsto d = 23 - 13.$

$\large\red{\mapsto\boxed{ \blue{ \bf d = 10.}}}$

$\bf\therefore$ The Required Answer Of (a) is 10.

And,

In (b) we have to find the first three digit term i.e. first term which is greater than 99.

$\large\blue{\mapsto \boxed{ \red{ \bf a_n > 99.}}}$

So by the formula we get:-

$\tt\mapsto a + (n - 1)d > 99.$

$\tt\mapsto13 + (n - 1)10 > 99.$

$\tt\mapsto13 + 10n - 10 > 99.$

$\tt\mapsto3 + 10n > 99.$

$\tt\mapsto10n > 99 - 3.$

$\tt\mapsto10n > 96.$

$\tt\mapsto n > \dfrac{96}{10} .$

$\large\blue{\mapsto \boxed{ \red{ \bf n > 9.6}}}$

Also we know that n is a number of term so it can not be in decimal.

So The nearest non decimal number which is greater than 9.6 is 10.

And hence the 10th term of the Sequence is first three digit number.

Therefore,

$\tt\mapsto a_{10} = a + (10 - 1)d.$

$\tt\mapsto a_{10} = 13 + (10 - 1)10.$

$\tt\mapsto a_{10} = 13 + (9)10.$

$\tt\mapsto a_{10} = 13 + 90.$

$\large \red{\mapsto \boxed{ \blue{ \bf a_{10 = 103.}}}}$

$\bf\therefore$ The First Three Digit Number Of The Sequence Is 103.

And hence the answer of (b) is 103.

Answer 2

Answer:

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