Question

18th term of the A.p,rout 2 rout 3,2,rout 5,2 .....

Answer 1

$\implies\sf 35 \sqrt{2}$

Given:-

• AP:- $\sf \sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2},.....a_{n}$

To find:-

• 18th term of AP

Solution:-

• First term (a) = 2

• Common difference (d) = $\sf a_{2} - a_{1} = 3 \sqrt{2} - \sqrt{2} = 2 \sqrt{2}$

★ Let's $\sf a_{18}$ be $\sf a_{n}$ term So,

$\sf a_{18} = \sqrt{2} + (18 - 1)2 \sqrt{2}$

$:\implies\sf \sqrt{2} + 17 \times 2 \sqrt{2}$

$:\implies\sf \sqrt{2} + 34 \sqrt{2}$

$:\implies{\underline{\underline{\sf{\red{35 \sqrt{2}}}}}}$

$\rule{200}{3}$

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Question}}}}}}}}$

Find the 18th term of AP $\sqrt 2 , 2 \sqrt 3 , 2 \sqrt 5 , .........$

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Solution}}}}}}}}$

${\bold{\blue{\boxed{\bf{Given}}}}}$

• first term = a = √2
• common diff. = d = $2 \sqrt 2$
• no. of times = n = 18

${\bold{\blue{\boxed{\bf{Given}}}}}$

$a_{18} = ???$

${\bold{\blue{\boxed{\bf{Formulae \: used}}}}}$

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

${\bold{\blue{\boxed{\bf{solution}}}}}$

$a_{18} =\sqrt 2 + (18 - 1 ) 2\sqrt 2$

$a_{18} = \sqrt 2 + 17 \times 2\sqrt 2$

$a_{18} =\sqrt 2 + 34 \sqrt 2$

$a_{18}= 35 \sqrt 2$

${\bold{\blue{\boxed{\boxed{\bf{35\sqrt 2}}}}}}$

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