Question

sum of first 10 term is 100 and sum of 10 term is 300 find the AP

Answer 1

Answer:

A.P

19, 21, 23, 25, 27, 29, 31 ...

Step-by-step explanation:

sₙ = 100

n = 10

d = d

a = a

100 = $\frac{10}{2}$  [ 2a - d(10-1)]

100 = 5[ 2a - d(9)]

20 = 2a - 9d                   => Transpose 100÷5 = 20

2a = 20 + 9d (eq.1)  

As after 10th term, it starts from 11th term

a  = a + 10d

d  = d

sₙ= 300

n = 10

300 = $\frac{10}{2}$ [2(a+10d) - d(10-1)]

300 = 5 [2a + 20d - 9d]             =>  2(a+10d) = 2a + 20d

60 = 2a + 11d                    => Transpose 300÷5 = 60

2a = 60-11d (eq.2)

Equating eq.1 and eq.2

20 + 9d = 60-11d

11d + 9d = 60 - 20                    => Transpose

20d = 40

d = 2

Substitute d in eq.1

2a = 20 + 18                             => 9d = 9 x 2 = 18

2a = 38

 a = 19                                       => transpose

a = 19

d = 2

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