Question

In an AP's if a=3. ,
d=3,an=60 find n and sn

Answer 1

Question :–

▪︎ In an A.P. if a = 3 , d = 3 , ${ \bold{ a_{n} = 60 }}$ then find n and ${ \bold{ s_{n} }}$.

ANSWER :–

$\\ \longrightarrow{ \bold{ n = 20 }}$

$\\ \longrightarrow{ \bold{ s_{n} = 630 }}$

EXPLANATION :–

GIVEN :–

• First term of A.P.(a) = 3

• Common difference (d) = 3

• nth term (${ \bold{ a_{n} }}$) = 60

TO FIND :–

• Total number of term (n) = ?

• Sum of n term ( ${ \bold{ s_{n} }}$ ) = ?

SOLUTION :–

☞ nth term of A.P. –

$\\ \implies{ \red {\boxed{ \bold{ a_{n} = a + (n - 1)d}}}}$

• Now put the values –

$\\ \implies{ \bold{ 60 = 3 + (n - 1)3}}$

$\\ \implies{ \bold{ 60 = \cancel3 + 3n - \cancel3}}$

$\\ \implies{ \bold{ 60 = 3n }}$

$\\ \implies{ \boxed{ \bold{ n = 20 }}}$

☞ Sum of n term –

$\\ \implies{ \red { \boxed{ \bold{ s_{n} = \dfrac{n}{2}[2a + (n - 1)d] }}}}$

• Now put the values –

$\\ \implies{ \bold{ s_{n} = \cancel \dfrac{20}{2}[2(3) + (20 - 1)3] }}$

$\\ \implies{ \bold{ s_{n} = 10[6 +19 \times 3] }}$

$\\ \implies{ \bold{ s_{n} = 10[6 +57] }}$

$\\ \implies{ \bold{ s_{n} = 10(63) }}$

$\\ \implies{ \boxed{ \bold{ s_{n} = 630 }}}$

Answer 2

Given: In an AP,

• a [first term] = 3
• d [common difference] = 3
• a_n [last term] = 60

To find: n [the number of terms] and S_n [sum of the terms.]

Answer:

Let's first find n.

$\sf The\ general\ form\ of\ a\ term\ is\ \bf a_n\ =\ a\ +\ (n\ -\ 1)d.$

Substituting the values,

$\sf 60\ =\ 3\ +\ (n\ -\ 1)3\\\\\\60\ =\ 3\ +\ 3n\ -\ 3\\\\\\60\ =\ 3n\\\\\\\dfrac{60}{3}\ =\ n\\\\\\20\ =\ n$

Therefore, the number of terms is 20.

Now, let's find the sum of the AP.

$\bf S_n\ =\ \dfrac{n}{2}\Bigg[2a\ +\ (n\ -\ 1)d\Bigg]$

Substituting the values,

$\sf S_n\ =\ \dfrac{20}{2}\Bigg[2\ \times\ 3\ +\ (20\ -\ 1)3\Bigg]\\\\\\S_n\ =\ 10\Bigg[6\ +\ 19\ \times 3\Bigg]\\\\\\S_n\ =\ 10\Bigg[6\ +\ 57\Bigg]\\\\\\S_n\ =\ 10\ \times\ 63\\\\\\S_n\ =\ 630$

Therefore, the AP has 20 terms and its sum is 630.