Question

The sun of first 20 terms of an AP is 400 and that of 40 terms is 1600. Find the sum of first 10 terms and that of n terms.

Answer 1

Sn = n/2 [2a+ (n-1)d]
S20 = 20/2 [2a +(20-1)d] =400
10[2a+(20-1)d]=400
[2a+19d]=40 -----(1)

S40= 40/2[2a+ (40-1)d] = 1600
20[2a+(40-1)d] = 1600
[2a+ (39d] = 1600/20 = 80-----(2)
now after eliminating we get.
2a + 39d = 80
2a + 19d = 40
changing the signs
20 d =40
d= 2
putting value of d in eq (1)
2a + 19×2=40
2a +38=40
2a =2
a=1
S 10 = 10/2[2a + 9d]
5[2+ 9(2)]= 10+ 90= 100

Answer 2

Hey there !!



Given :-

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


And,

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


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,

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


✔✔ Hence, it is solved ✅✅.



THANKS


#BeBrainly.