Question

find S3 for an A.P 3,5,7,9

​

Answer 1

Answer:

n=3

a=3

d=5-3=2

S3 = n/2[2a+(n-1)d]

= 3/2[2*3+(3-1)2]

= 3/2[6+4]

= 3/2*10

=3*5

= 15

Answer 2

$\red{\bigstar \large\underline{\boxed{\bf\pink{Question:-}}}}$

$\tt Find\: S_3\: for\: an\: A.P\: 3,5,7,9$

$\red{\bigstar \large\underline{\boxed{\bf\pink{Answer:-}}}}$

$\bf \green{\dag{\underline {Given:-}}}$

➠${\orange {\sf A.P\: is \:3,5,7,9}}$

$\: \: \: \: \: \: \: \: \: ➪{\orange {\sf a=3}}\\ \: \: \: \: \: \: \: \: \: ➪{\orange{\sf d=2}}\\ \: \: \: \: \: \: \: \: \: \: ➪{\orange{\sf n=3}}$

$\bf \green{\dag{\underline {To\:Find:-}}}$

➠${\orange {\sf Sum\:of \:3\:terms \:of\:an\:A.P}}$

$\bf \green{\dag{\underline {Solution:-}}}$

${\pink {\boxed {\red{\pmb {S_n=\dfrac{n}{2}{\Big[2a+(n-1)d\Big]}}}}}}$

$\longrightarrow{\purple {\sf S_3=\dfrac{3}{2}\Big[2\times 3+(3-1)2\Big]}}$

$\longrightarrow{\purple {\sf S_3=\dfrac{3}{2}\Big[6+2\times 2\Big]}}$

$\longrightarrow{\purple {\sf S_3=\dfrac{3}{2}\Big[6+4\Big]}}$

$\longrightarrow{\purple {\sf S_3=\dfrac{3}{2}\Big[10\Big]}}$

$\longrightarrow{\purple {\sf S_3=3 \times 5}}$

$\longrightarrow{\underbrace {\overbrace {\color{yellow} {\boxed {\mathfrak {S_3=15}}}}}}$

${\pink {\underline {\boxed {\sf \therefore The\:sum\:of\:3\:terms\:in\:an\:A.P\:is\:15}}}}$