Question

The 11th term of the AP = √2, 3√2, 5√2, ...... is *​

Answer 1

Answer:

$\huge{\underline{\underline{\tt{\redſ.Question}}}}$

The 11th term of the AP = √2, 3√2, 5√2, ...... is 

$\huge{\underline{\underline{\tt{\blueſ.Answer}}}}$

$\huge{\underline{\underline{\tt{\pinkſ.Given\ series\ is:}}}}$

√2,3√2,5√2,.....

We know that tn=a+(n-1)d

Thus a=√2, d=2√2

$\huge{\underline{\underline{\tt{\blueſ.Therefore}}}}$

t11=√2+(11-1)2√2

=√2+(10)2√2

=√2+20√2

=21√2

Therefore 11th term is

$\huge{\underline{\underline{\tt{\greenſ.=21√2}}}}$

Answer 2

$\bf \large{ \underline{ \underline{ \pink{Given : }}}}$

$\bf \small{AP} \sf \small {: \sqrt{2} ,\: 3 \sqrt{2}, \: 5 \sqrt{2}... }$

$\bf \large { \underline { \underline{ \red{Required \: answer : }}}}$

$\bf \small{The \: {11}^{th} term \: of \: the \: given \: AP.}$

$\bf \large { \underline{ \underline{ \green{Solution : }}}}$

We have,

$\bf \small{a} \sf \small{ = \sqrt{2} }$

$\bf \small{d} \sf \small{ = 3 \sqrt{2} - \sqrt{2} = 2 \sqrt{2} }$

$\bf \small{n = } \sf \small{11}$

$\bf \small{a11 = } \sf \small{a + 10d = \sqrt{2} + 10 \sqrt{2} }$

$\implies \bf \small{a11 = } \sf \small{11 \sqrt{2} }$

$\bf \large { \underline{ \underline{ \blue{Formula \: used : }}}}$

$\bf \small{a_n = } \sf \small{a + (n - 1)d}$