Question

the sum of first 20 terms of an AP is 400 and the sum of first 40 terms is 1600 . find s10

Answer 1

S 20 =400
a + 19d=400_____1
S 40 =1600
a+39d=1600____2
sub.eq1 by 2
a+39d=1600
a+19d=400
-. -. -
20d=1200
d= 1200÷20
d=60

then you can put value of d in eq1 ......you will get value of a
......and then put value of a&d in term ....a1 to a10....

Answer 2

Answer:100.

Step-by-step explanation:

Hey there !!

Given :-

$\begin{lgathered}S_{20} = 400. \\ \\ = > \frac{n}{2} (2a + (n - 1)d) = 400. \\ \\ = > \frac{20}{2} (2a + (20 - 1)d) = 400. \\ \\ = > 10(2a + 19d) =4 00. \\ \\ = > 2a + 19d = \frac{400}{10} . \\ \\ = > 2a + 19d = 40..........(1).\end{lgathered}$

And,

$\begin{lgathered}S_{40} = 1600. \\ \\ = > \frac{n}{2} (2a + (n - 1)d) = 1600. \\ \\ = > \frac{40}{2} (2a + (40 - 1)d) = 1600. \\ \\ = > 20(2a + 39d) =16 00. \\ \\ = > 2a + 39d = \frac{1600}{20} . \\ \\ = > 2a + 39d = 80..........(2).\end{lgathered}$

​

Substracting in equation (2) and (1) , we get

2a + 39d = 80 .

2a + 19d = 40 .

(-).....(-).......(-).

___________

=> 20d = 40 .

=> d = 40/20 .

•°• d = 2 .

On putting the value of d in equation (1), we get

=> 2a + 19(2) = 40.

=> 2a = 40 - 38 .

=> 2a = 2 .

=> a = 2/2 .

•°• a = 1 .

Now,

$\begin{lgathered}S_{10} = \frac{n}{2} (2a + (n - 1)d). \\ \\ = \frac{10}{2} (2 \times 1 + (10 - 1)2). \\ \\ = 5(2 + 18). \\ \\ = 5 \times 20. \\ \\ \huge \boxed{ \green{ = 100.}}\end{lgathered}$

​

​

✔✔ Hence, it is solved ✅✅.

THANKS

#BeBrainly.