Question

Find the sum of first 12 terms of the arithmetic progression 30, 39, 48......

Answer 1

Question

Find the sum of first 12 terms of the arithmetic progression 30, 39, 48......

Answer

First find the 12th term :-

$\mathtt{a_{12} = a_{1} + (n - 1)d}$

$\mathtt{a _{12} = 30 + 11(9)}$

$\mathtt{a _{12_{term}} = 129}$

Now, let's find the sum

$\mathtt{s_{n} = \frac{n(a_{1} + a _{n}) }{2}}$

$\mathtt{s_{12} = \frac{12(30 + 129)}{2}}$

$\mathtt{s_{12} = \frac{1908}{2}}$

$\mathtt \red{s_{12} = 984}$

Therefore, the sum of 12th term is 984.

Important

† The general form of AP is a,a+d,a+2d,a+3d,and so on

† $\mathtt{S_{n}=\frac{n(a_{1}+a_{n}}{2}}$

† $\mathtt{a_{n}=a_{1}+(n-1)}$

Answer 2

${\huge{\rm{\underline{\underline{Question:-}}}}}$

Q) Find the sum of first 12 terms of the Airihmetic Progression 30 , 39 , 48 , ...

${\huge{\underline{\underline{\rm{Answer\checkmark}}}}}$

★ Given :–

• A.P. → 30 , 39 , 48 , ....

★ To Find :–

• Sum of first 12 terms ( $s_{12}$ )

★ Solution :–

• First term , a = 30
• Common difference , d = 39 - 30 = 9
• number of terms , n = 12

Using the Formula to find sum :

$\boxed{ \sf{s _{n} = \frac{n}{2} \{2a + (n - 1)d \}}}$

Just put the values here ..

$\mapsto \rm \: s _{12} = \frac{ \cancel{12}}{ \cancel2} \{2 \times 30 + (12 - 1)9 \}$

$\mapsto \rm \: s _{12} = 6 \{60 + 11 \times 9 \}$

$\mapsto \rm \: s _ {12} = 6 \{60 + 99 \}$

$\mapsto \rm \: s _{12} = 6 \times 159$

$\mapsto \underline{ \boxed{ \rm{ \pink{s _{12} = 954}}}}$

So ,

The sum of first 12 terms of this A.P. is 954 .

_________________________

${\large{\underline{\rm{★\;Know\;More}}}}$

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

• ${\sf{a_n=a+(n-1)d}}$

• ${\sf{S_n=\dfrac{n}{2}\{2a+(n-1)d\}}}$

• ${\sf{S_n=\dfrac{n}{2}\{a+a_n\}}}$

Here ,

$\purple{\ddot{ \smile}} \: \tt \: a = first \: term \: .$

$\green{ \ddot{ \smile}} \: \tt \: d = common \: difference \: .$

$\pink{ \ddot{ \smile}} \: \tt \: n = no. \: of \: terms \: .$

$\blue{ \ddot{ \smile}} \: \tt \: a _{n} = {n}^{th} \: term$

$\orange{ \ddot{ \smile}} \: \tt \: s _{n} = sum \: of \: n \: terms$