Question

The sum of the first 10 terms of an arithmetic progression is 120 and the sum of first twenty is 840. find sum of first 30 terms​

Answer 1

Answer:

The sum of first 30 terms of the arithmetic progression is 2160.

Explanation:

For an arithmetic progression, the sum of first $n$ terms with first term as $a$ and common difference $d$ is given as:

$S_n=\frac{n}{2}(2a+(n-1)d)$

Now, it is given that:

$For\ n=10,S_n=120\\For\ n=20,S_n=840$

Now, plug in these values and frame two equations in $a\ and\ d$

$S_{10}=\frac{10}{2}(2a+(10-1)d)\\120=5(2a+9d)\\2a+9d=\frac{120}{5}\\2a+9d=24------------1$

$S_{20}=\frac{20}{2}(2a+(20-1)d)\\840=10(2a+19d)\\2a+19d=\frac{840}{10}\\2a+19d=84-----------2$

Now, we solve equations (1) and (2) for $a\ and\ d$. Subtract equation (1) from equation (2). This gives,

$2a+19d-2a-9d=84-24\\19d-9d=60\\10d=60\\d=\frac{60}{10}=6$

Now, plug in the value of $d=6$ in equation (1) and solve for $a$.

$2a+9(6)=24\\2a+54=24\\2a=24-54\\2a=-30\\a=\frac{-30}{2}=-15$

Plug in the values of $a=-15,\ n=30\ and\ d=6$ in the sum formula to find the sum of first 30 terms.

Now, the sum of first 30 terms is given as:

$S_{30}=\frac{30}{2}(2(-15)+(30-1)(6))\\S_{30}=15(-30+29(6))\\S_{30}=15(-30+174)\\S_{30}=15(144)=2160$

Therefore, the sum of first 30 terms of the arithmetic progression is 2160.