Question

1) The 10th term or the sequence.
root3, root 12, root127, ... is .​

Answer 1

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Q) Find the 10th term of the sequence :

$\sqrt{3} \: , \: \sqrt{12} \: , \: \sqrt{27} \: , \: ...$

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• The given sequence is an Arithmetic Progression.

• Arithmetic Progression : It is a sequence in which adjacent terms differ with common difference.

${\textit{\textbf{Finding~Common~Difference:}}}$

We have A.P.

$\sqrt{3} \: , \: \sqrt{12} \: , \: \sqrt{27} \:, \: ...$

Modify it..

$\rightarrow \: \sqrt{3} \: , \: \sqrt{4 \times 3} \: , \: \sqrt{9 \times 3} \: , \: ...$

$\rightarrow \: \sqrt{3} \: , \: 2 \sqrt{3} \: , \: 3 \sqrt{3} \: , \: ...$

$\colon \implies \sf \: common \: diffference (d)= 2 \sqrt{3} - \sqrt{3} \: or \: 3 \sqrt{3} - 2 \sqrt{3} \\ \\ \colon \implies \boxed{\sf d = {\bf {\pink{\sqrt{3} }}}}$

${\textit{\textbf{Finding~The~10{th}~term}}}$

In the given AP,

• First Term = √3 .

$\star \: \boxed{ \sf{n {}^{th} \: term = first \: term + (n - 1) \times common \: difference}}$

$\colon \implies \sf \: 10 {}^{th} \: term = first \: term + (10 - 1) \times common \: difference \\ \\ \colon \implies \sf \: a _{10} = \sqrt{3} + (10 - 1) \times \sqrt{3} \\ \\ \colon \implies \sf \: a _{10} = \sqrt{3} + 9 \sqrt{3} \\ \\ \colon \implies \underline{ \boxed{ \bf{ \purple{a _{10} = 10 \sqrt{ 3}}}} }$

So,

• The 10th term of this A. P. would be 10√3 .

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★ Some A. P. related formulas :

$\bull\rm\:n\:th\:term=a+(n-1) d$

$\bull \rm \: S _{n} = \frac{n}{2} \{2a + (n - 1)d \} \:$

$\bull \rm \: S_{n} = \frac{n}{2} \{first \: term + last \: term \} \:$

Here,

• n = number of terms.
• a = first term .
• d = common difference.
• $\rm S_n$ = sum of first n terms .

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Answer 2

Answer:

is u r answer hope this helps u