Question

in an ap 18th term is 180and 36th term is 360find the sum of first 53terms​

Answer 1

Step-by-step explanation:

hand helping is a thing

Answer 2

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

In an A.P. 18th term is 180 and 36th term is 360.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The sum of first 53 terms.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We know that formula of an A.P;

$\boxed{\bf{a_{n}=a+(n-1)d}}}}$

• a is the first term.
• d is the common difference.
• n is the term of an A.P.

A/q

$\mapsto\sf{a_{18}=180}\\\\\mapsto\sf{a+(18-1)d=180}\\\\\mapsto\sf{a+17d=180.......................(1)}$

&

$\mapsto\sf{a_{36}=360}\\\\\mapsto\sf{a+(36-1)d=360}\\\\\mapsto\sf{a+35d=360.......................(2)}$

Subtracting equation (1) and equation (2),we get;

$\longrightarrow\sf{\cancel{a}+17d \cancel{-a}-35d=180-360}\\\\\longrightarrow\sf{17d-35d=-180}\\\\\longrightarrow\sf{-18d=-180}\\\\\longrightarrow\sf{d=\cancel{\dfrac{-180}{-18}}} \\\\\longrightarrow\sf{\orange{d=10}}$

Putting the value of d in equation (1),we get;

$\longrightarrow\sf{a+17d=180}\\\\\longrightarrow\sf{a+17(10)=180}\\\\\longrightarrow\sf{a+170=180}\\\\\longrightarrow\sf{a=180-170}\\\\\longrightarrow\sf{\orange{a=10}}$

Now;

We know that formula of the sum of an A.P;

$\boxed{\bf{S_{n}=\frac{n}{2} \big[2a+(n-1)d\big]}}}}$

$\longrightarrow\sf{S_{53}=\dfrac{53}{2} \bigg[2(10)+(53-1)(10)\bigg]}\\\\\\\longrightarrow\sf{S_{53}=\dfrac{53}{2} \bigg[20+52\times 10\bigg]}\\\\\\\longrightarrow\sf{S_{53}=\dfrac{53}{2} \bigg[20+520\bigg]}\\\\\\\longrightarrow\sf{S_{53}=\dfrac{53}{\cancel{2}} \times \cancel{540}}\\\\\\\longrightarrow\sf{S_{53}=53\times 270}\\\\\\\longrightarrow\sf{\orange{S_{53}=14310}}$

Thus;

The sum of first 53 terms is 14310.