Question

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Answer 1

Answer:

q8

let first term =a =12and common difference -d in an AP

nth term =an =a+(n-1)d ---(1)

7th term=a7=a+6d-----(2)

a11-a+10d----(3)

given

a11-a7=24

a+10d-(a+6d)=24 a+10d-a-6d=24

4d=24

d=6--(4)

required term =a20

=a+19d

=12+19*6

=12+114

=126

q9

Let a and d respectively be the first term and common difference of the AP.

Given a9 = 0

So, a + (9-1)d=0

a+8d=0

a= -8d

Now, 29th term=a+28d

=-8d+28d

= 20d = 2 x 10d

= 2(-8d + 18d)

=2(a+18d)

= 2 x 19th term

Thus, the 29th term of the AP is twice the 19th term.

q10

a=7

d=3

therefore, ATQ; An=55

a+(n-1)d=55

7+(n-1)3=55

(n-1)3-55-7

(n-1)=48/3

n-1-16

=>n=17

therefore, the 17th term is 55

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